To begin, let us start with the $\displaystyle 10\Omega$ resistor and the $\displaystyle 15\Omega$ resistor that are in parallel. In parallel we can add resistors through the equation $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{10} + \frac {1}{15}$

$\displaystyle \frac{1}{R_{eq}} = \frac{3}{30} + \frac{2}{30}$

$\displaystyle \frac{1}{R_{eq}} = \frac{5}{30}$

$\displaystyle R_eq = \frac{30}{5}$

$\displaystyle R_{eq} = 6$

This new resistor is now in series with the two $\displaystyle 4\Omega$ resistors.  In series we can just add these resistors up.

$\displaystyle R_{eq} = 4 + 4 + 6$

$\displaystyle R_{eq} = 14$

This new resistor is now in parallel with the $\displaystyle 14\Omega$ resistor. In parallel we can add resistors through the equation  $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_eq} = \frac{1}{14} +\frac{1}{14}$

$\displaystyle \frac{1}{R_eq} =\frac{2}{14}$

$\displaystyle R_{eq} = \frac{14}{2}$

$\displaystyle R_{eq} = 7$

This new resistor is now in series with the $\displaystyle 3\Omega$ and $\displaystyle 5\Omega$ resistor.  In series we can just add these resistors up.

$\displaystyle R_{eq} = 7 + 3 + 5$

$\displaystyle R_{eq} = 15$

Now that we have the equivalent resistance of the circuit.  We can now determine the current flowing out of the battery.

$\displaystyle V=IR$

Rearrange to solve for current.

$\displaystyle I = \frac{V}{R}$

$\displaystyle I = \frac{60V}{15\Omega}$

$\displaystyle I = 4A$

One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.

We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series.  So we can put this information into our chart.

Using Ohm’s Law we can determine the voltage for each of these two resistors.

$\displaystyle V=IR$

$\displaystyle V = (4A)(5\Omega)$

$\displaystyle V = 20V$

$\displaystyle V = (4A)(3\Omega)$

$\displaystyle V = 12V$

$\displaystyle V_{battery} - V_{5\Omega} - V_{3\Omega} - V_{14\Omega} = 0V$

$\displaystyle 60V - 20V - 12V - V_{14\Omega} = 0V$

$\displaystyle 28V - V_{14\Omega} = 0V$

$\displaystyle 28V = V_{14\Omega}$

We can now use Ohm’s law and our voltage to determine the current going through the 14 Ohm resistor.

$\displaystyle V = IR$

$\displaystyle I = \frac{V}{R}$

$\displaystyle I = \frac{14V}{14\Omega}$

$\displaystyle I = 1A$

Next we can analyze the current in the Junction between the 5, 14 and 4 Ohm resistors.  Kirchoff’s laws state that the sum of the current flowing in and out of junction must equal 0.

We have $\displaystyle 4A$ flowing in from the $\displaystyle 5 \: Ohm$ resistor and $\displaystyle 1A$ flowing out to go through the $\displaystyle 14 \: Ohm$ resistor.

$\displaystyle 4A - 2A - I_{4\Omega} = 0$

$\displaystyle 2A -I_{4\Omega} = 0$

$\displaystyle I_{4\Omega} = 2A$

So the current going through the 4 Ohm resistor is 3 amps.  This will be the same for both 4 ohm resistors as both have the same current going into and out of the junctions near them.  We can then use this to determine the voltage drop across each of the resistors.

$\displaystyle V = IR$

$\displaystyle V = (2A)(4\Omega)$

$\displaystyle V = 8V$

Now let’s add this to our chart.

We can now use Kirchoff’s loop law to determine the voltage across the 10 Ohm resistor.  Let’s analyze the loop that goes from the 5, to the 4 to the 10 back through the 4 and then through 3 Ohm resistor.

$\displaystyle V_{battery} - V_{5\Omega} - V_{4\Omega} - V_{10\Omega} - V_{4\Omega} - V_{3\Omega} = 0V$

$\displaystyle 60V - 20V - 8V - V_{10\Omega} - 8V - 12V = 0V$

$\displaystyle 12V - V_{10\Omega} = 0V$

$\displaystyle V_{10|Omega} = 12V$

We can now use Ohm’s Law to determine the current through the 10 Ohm resistor.

$\displaystyle V = IR$

$\displaystyle I = \frac{V}{R}$

$\displaystyle I = \frac{12V}{10\Omega}$

$\displaystyle I = 1.2A$

To begin, let us start with the $\displaystyle 10\Omega$ resistor and the $\displaystyle 15\Omega$ resistor that are in parallel. In parallel we can add resistors through the equation $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{10} + \frac {1}{15}$

$\displaystyle \frac{1}{R_{eq}} = \frac{3}{30} + \frac{2}{30}$

$\displaystyle \frac{1}{R_{eq}} = \frac{5}{30}$

$\displaystyle R_eq = \frac{30}{5}$

$\displaystyle R_{eq} = 6$

This new resistor is now in series with the two $\displaystyle 4\Omega$ resistors.  In series we can just add these resistors up.

$\displaystyle R_{eq} = 4 + 4 + 6$

$\displaystyle R_{eq} = 14$

This new resistor is now in parallel with the $\displaystyle 14\Omega$ resistor. In parallel we can add resistors through the equation $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_eq} = \frac{1}{14} +\frac{1}{14}$

$\displaystyle \frac{1}{R_eq} =\frac{2}{14}$

$\displaystyle R_{eq} = \frac{14}{2}$

$\displaystyle R_{eq} = 7$

This new resistor is now in series with the $\displaystyle 3\Omega$ and $\displaystyle 5\Omega$ resistor.  In series we can just add these resistors up.

$\displaystyle R_{eq} = 7 + 3 + 5$

$\displaystyle R_{eq} = 15$

Now that we have the equivalent resistance of the circuit.  We can now determine the current flowing out of the battery.

$\displaystyle V=IR$

Rearrange to solve for current.

$\displaystyle I = \frac{V}{R}$

$\displaystyle I = \frac{60V}{15\Omega}$

$\displaystyle I = 4A$

One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.

We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series.  So we can put this information into our chart.

Using Ohm’s Law we can determine the voltage for each of these two resistors.

$\displaystyle V=IR$

$\displaystyle V = (4A)(5\Omega)$

$\displaystyle V = 20V$

$\displaystyle V = (4A)(3\Omega)$

$\displaystyle V = 12V$

We can now use Kirchoff’s loop law through the loop of the 5, 3, and 14 Ohm resistor to determine the voltage that is traveling through the 14 Ohm resistor.

$\displaystyle V_{battery} - V_{5\Omega} - V_{3\Omega} - V_{14\Omega} = 0V$

$\displaystyle 60V - 20V - 12V - V_{14\Omega} = 0V$

$\displaystyle 28V - V_{14\Omega} = 0V$

We can now use Ohm’s law and our voltage to determine the current going through the 14 Ohm resistor.

$\displaystyle V = IR$

$\displaystyle I = \frac{V}{R}$

$\displaystyle I = \frac{14V}{14\Omega}$

$\displaystyle I = 1A$

We can now add this information to our chart.

Next we can analyze the current in the Junction between the 5, 14 and 4 Ohm resistors.  Kirchoff’s laws state that the sum of the current flowing in and out of the junction must equal 0.

We have 4A flowing in from the 5 Ohm resistor and 1A flowing out to go through the 14 Ohm resistor.

$\displaystyle 4A - 2A - I_{4\Omega} = 0$

$\displaystyle 2A -I_{4\Omega} = 0$

$\displaystyle I_{4\Omega} = 2A$

So the current going through the 4 Ohm resistor is 3 amps.  This will be the same for both 4 ohm resistors as both have the same current going into and out of the junctions near them.  We can then use this to determine the voltage drop across each of the resistors.

$\displaystyle V = IR$

$\displaystyle V = (2A)(4\Omega)$

$\displaystyle V = 8V$

Now let’s add this to our chart.

We can now use Kirchoff’s loop law to determine the voltage across the 10 Ohm resistor.  Let’s analyze the loop that goes from the 5, to the 4 to the 10 back through the 4 and then through 3 Ohm resistor.

$\displaystyle V_{battery} - V_{5\Omega} - V_{4\Omega} - V_{10\Omega} - V_{4\Omega} - V_{3\Omega} = 0V$

$\displaystyle 60V - 20V - 8V - V_{10\Omega} - 8V - 12V = 0V$

$\displaystyle 12V - V_{10\Omega} = 0V$

$\displaystyle V_{10\Omega} = 12V$

We can now use Ohm’s Law to determine the current through the 10 Ohm resistor.

$\displaystyle V = IR$

$\displaystyle I = \frac{V}{R}$

$\displaystyle I = \frac{12V}{10\Omega}$

$\displaystyle I = 1.2A$

We can now add this information to our chart.

We can now analyze the junction between the 4, 10 and 15 Ohm resistor.

Kirchoff’s laws state that the sum of the current flowing in and out of the junction must equal 0.

We have 2A flowing in from the 4 Ohm resistor and 1.2A flowing out to go through the 10 Ohm resistor.

$\displaystyle 2A - 1.2A - I_{15\Omega} = 0$

$\displaystyle 0.8A -I_{15\Omega} = 0$

$\displaystyle I_{15\Omega} = 0.8A$

To begin we need to simplify the circuit to get the equivalent resistance.  Let’s start with the 3 and 6 Ohm resistors in parallel.

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{2}{6} + \frac{1}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{3}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{2}$

$\displaystyle R_{eq} = 2\Omega$

Now we can add this resistor to the 4 Ohm resistor as they are in series.

$\displaystyle R_{eq} = R_1 + R_2 +...$

$\displaystyle R_{eq} = 4 + 2$

$\displaystyle R_{eq} = 6\Omega$

We can now determine the current coming out of the battery using Ohm’s Law.

$\displaystyle V=IR$

Rearrange to solve for current.

$\displaystyle I=\frac{V}{R}$

$\displaystyle I=\frac{6V}{6\Omega}$

$\displaystyle I = 1A$

The current coming out of the battery will be the same current that moves through the 4 Ohm resistor.  So we can determine the voltage drop across the 4 ohm resistor.

$\displaystyle V = IR$

$\displaystyle V = (1A)(4\Omega)$

$\displaystyle V = 4V$

We can then use Kirchoff’s loop law to determine the voltage drop across the 6 Ohm resistor.  Let’s analyze the loop that goes from the battery to the 4 ohm resistor and through the 6 ohm resistor.

$\displaystyle V_{battery} - V_{4\Omega} - V_{6\Omega} = 0V$

$\displaystyle 6V - 4V - V_{6\Omega} = 0V$

$\displaystyle 2V -V_{6\Omega} = 0V$

$\displaystyle V_{6\Omega} = 2V$

To begin we need to simplify the circuit to get the equivalent resistance.  Let’s start with the 3 and 6 Ohm resistors in parallel.

 $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{2}{6} + \frac{1}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{3}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{2}$

$\displaystyle R_{eq} = 2\Omega$

Now we can add this resistor to the 4 Ohm resistor as they are in series.

$\displaystyle R_{eq} = R_1 + R_2 +...$

$\displaystyle R_{eq} = 4 + 2$

$\displaystyle R_{eq} = 6\Omega$

We can now determine the current coming out of the battery using Ohm’s Law.

$\displaystyle V=IR$

Rearrange to solve for current.

$\displaystyle I=\frac{V}{R}$

$\displaystyle I=\frac{6V}{6\Omega}$

$\displaystyle I = 1A$

The current coming out of the battery will be the same current that moves through the 4 Ohm resistor.  So we can determine the voltage drop across the 4 ohm resistor.

$\displaystyle V = IR$

$\displaystyle V = (1A)(4\Omega)$

$\displaystyle V = 4V$

Kirchoff’s loop rules states the sum of the current going into and out of the junction must equal 0. In other words, the current going in must equal the current going on.  Current is a measure of the flow of charge.  Therefore, this law is conservation of charge as the number of electrons going into a junction must equal the number of electrons flowing out.

To begin, let us start with the $\displaystyle 10\Omega$ resistor and the $\displaystyle 15\Omega$ resistor that are in parallel. In parallel we can add resistors through the equation $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{10} + \frac {1}{15}$

$\displaystyle \frac{1}{R_{eq}} = \frac{3}{30} + \frac{2}{30}$

$\displaystyle \frac{1}{R_{eq}} = \frac{5}{30}$

$\displaystyle R_eq = \frac{30}{5}$

$\displaystyle R_{eq} = 6$

This new resistor is now in series with the two 4Ω resistors.  In series we can just add these resistors up.

$\displaystyle R_{eq} = 4 + 4 + 6$

$\displaystyle R_{eq} = 14$

This new resistor is now in parallel with the $\displaystyle 14\Omega$ resistor. In parallel we can add resistors through the equation $\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_eq} = \frac{1}{14} +\frac{1}{14}$

$\displaystyle \frac{1}{R_eq} =\frac{2}{14}$

$\displaystyle R_{eq} = \frac{14}{2}$

$\displaystyle R_{eq} = 7$

This new resistor is now in series with the $\displaystyle 3\Omega$ and $\displaystyle 5\Omega$ resistor.  In series we can just add these resistors up.

$\displaystyle R_{eq} = 7 + 3 + 5$

$\displaystyle R_{eq} = 15$

Now that we have the equivalent resistance of the circuit.  We can now determine the current flowing out of the battery.

$\displaystyle V=IR$

Rearrange to solve for current.

$\displaystyle I = \frac{V}{R}$

$\displaystyle I = \frac{60V}{15\Omega}$

$\displaystyle I = 4A$

One of the best ways to work through a problem like this is to create a V= IR chart for all the components of the circuit.

We know that the current that is flowing out of the battery is the current that is flowing through both by 5 and 3 Ohm resistors since all of these are in series.  So we can put this information into our chart.

Using Ohm’s Law we can determine the voltage for each of these two resistors.

$\displaystyle V=IR$

$\displaystyle V = (4A)(5\Omega)$

$\displaystyle V = 20V$

$\displaystyle V = (4A)(3\Omega)$

$\displaystyle V = 12V$

We can now use Kirchoff’s loop law through the loop of the 5, 3, and 14 Ohm resistor to determine the voltage that is traveling through the 14 Ohm resistor.

$\displaystyle V_{battery} - V_{5\Omega} - V_{3\Omega} - V_{14\Omega} = 0V$

$\displaystyle 60V - 20V - 12V - V_{14\Omega} = 0V$

$\displaystyle 28V - V_{14\Omega} = 0V$

Kirchoff’s loop law states that the sum of the voltage around a loop must equal zero.  In other words, the voltage that is being provided by the batteries in the circuit must equal the voltage being used by the objects in the circuit.  Voltage is a measure of the potential difference, or energy within the circuit.  In other words, the battery does a certain amount of work and provides energy to the circuit which is then used by all the parts of the circuit.  Therefore this is an example of conservation of energy.

To begin we need to simplify the circuit to get the equivalent resistance.  Let’s start with the 3 and 6 Ohm resistors in parallel.

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{2}{6} + \frac{1}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{3}{6}$

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{2}$

$\displaystyle R_{eq} = 2\Omega$

Now we can add this resistor to the 4 Ohm resistor as they are in series.

$\displaystyle R_{eq} = R_1 + R_2 +...$

$\displaystyle R_{eq} = 4 + 2$

$\displaystyle R_{eq} = 6\Omega$

We can now determine the current coming out of the battery using Ohm’s Law.

$\displaystyle V=IR$

Rearrange to solve for current.

$\displaystyle I=\frac{V}{R}$

$\displaystyle I=\frac{6V}{6\Omega}$

$\displaystyle I = 1A$

The current coming out of the battery will be the same current that moves through the 4 Ohm resistor.  So we can determine the voltage drop across the 4 ohm resistor.

$\displaystyle V = IR$

$\displaystyle V = (1A)(4\Omega)$

$\displaystyle V = 4V$

We can then use Kirchoff’s loop law to determine the voltage drop from point $\displaystyle X$ to point $\displaystyle Y$.

Let’s analyze the loop that goes from the battery to the 4 ohm resistor and through the 3 ohm resistor.

$\displaystyle V_{battery} - V_{4\Omega} - V_{XtoY} = 0V$

$\displaystyle 6V - 4V - V_{XtoY} = 0V$

$\displaystyle 2V -V_{XtoY} = 0V$

$\displaystyle V_{XtoY} = 2V$

To answer this question we must consider Kirchoff’s Loop Law.  This law states that the voltage around any loop must equal 0.  In this case there are two different loops at play.  To begin, let’s start on the left with the 2 Volt battery.

As we start with the 2 Volt battery, we then move into the 1 Ohm resistor with $\displaystyle I_1$ going through it.  Ohm’s law states that the voltage is equal to the current times the resistance.  Therefore the voltage through this circuit is $\displaystyle 1I_1$ Since this resistor is using the voltage this will be a negative voltage when we sum around the loop.

We will continue our loop through the middle of the circuit into the 4 Volt battery.  This battery is facing the opposite direction from our 2 Volt battery and therefore will be a negative when it comes to our equation.  

Next is  the 2 Ohm resistor with $\displaystyle I_2$ going through it.  According to Ohm’s law the voltage being used by this resistor is equal to $\displaystyle 2I_2$  

When summarizing all of these parts we get an equation that looks like $\displaystyle 2 - 1I_1 - 4 - 2I_2 = 0$which simplifies down to $\displaystyle -2 - 1I_1 - 2I_2 = 0$ This is one of the equations available to us and therefore there is no need to analyze any other loops.