Question Video: Using Kirchhoff’s Laws to Calculate Current in a Circuit | Nagwa Question Video: Using Kirchhoff’s Laws to Calculate Current in a Circuit | Nagwa

# Question Video: Using Kirchhoff’s Laws to Calculate Current in a Circuit Physics • Third Year of Secondary School

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Consider the given circuit. What is the value of 𝐼?

04:27

### Video Transcript

Consider the given circuit. What is the value of 𝐼? (A) Nine amps, (B) six amps, (C) 4.5 amps, (D) three amps.

In this question, we are presented with the following circuit, and we want to calculate the value of the current 𝐼 labeled in the circuit. To answer this question, we will use Kirchhoff’s laws. Recall that Kirchhoff’s first law states that the sum of the currents into a junction, or node, in the circuit must be the same as the sum of the currents out of the junction, or node.

Let’s label the circuit diagram with the currents in the circuit. We can see that the current 𝐼 comes into the node highlighted in blue, and the currents 𝐼 one and 𝐼 two go out of the node highlighted in blue. Therefore, by using Kirchhoff’s first law at this node, we find that 𝐼 equals 𝐼 one plus 𝐼 two. We are given the value of 𝐼 one. So if we can find out the value of the current 𝐼 two, then we will be able to calculate the value of 𝐼.

To find the value of 𝐼 two, we will use Kirchhoff’s second law. Recall that Kirchhoff’s second law states that the sum of the potential difference across each component in a loop is equal to zero. Let’s consider loop one, which we’ve highlighted in the circuit. We can label the potential difference across the five-ohm resistor as 𝑉 𝑅 one, and we can label the potential difference across the 10-ohm resistor as 𝑉 𝑅 two. Then, by applying Kirchhoff’s second law to loop one, we find that 𝑉 𝑅 one plus 𝑉 𝑅 two equals zero.

We can get expressions for 𝑉 𝑅 one and 𝑉 𝑅 two by using Ohm’s law. Recall that Ohm’s law can be written as 𝑉 equals 𝐼𝑅, where 𝑉 is the potential difference, 𝐼 is the current, and 𝑅 is the resistance. We also need to use some sign conventions when using Kirchhoff’s second law. When the loop passes through a resistor in the same direction as the current, the 𝐼𝑅 term is negative because the current goes in the direction of decreasing potential. When the loop passes through a resistor in the direction opposite to the current, the 𝐼𝑅 term is positive because this represents an increase in potential.

Now we can apply Ohm’s law. Let’s start with the five-ohm resistor. The current through the five-ohm resistor is the current 𝐼 one, which we’re told has a value of three amps. We also note that the loop passes through the resistor in the same direction as the current, so our 𝐼𝑅 term must be negative. When we apply Ohm’s law, we find that the potential difference across the five-ohm resistor 𝑉 𝑅 one is equal to minus three amps times five ohms equals minus 15 volts.

Next, let’s look at the 10-ohm resistor. The current through the 10-ohm resistor is the current 𝐼 two. We note this time that the loop passes through the resistor in the direction opposite to the current, so our 𝐼𝑅 term is positive. The potential difference across the 10-ohm resistor 𝑉 𝑅 two will be equal to 𝐼 two times 10 ohms. So by substituting the values of 𝑉 𝑅 one and 𝑉 𝑅 two into the equation we obtained from Kirchhoff’s second law for loop one, we have the equation minus 15 volts plus 10 ohms times 𝐼 two equals zero. Adding 15 volts to both sides of the equation, we get 10 ohms times 𝐼 two equals 15 volts. And by dividing both sides by 10 ohms, we find that 𝐼 two equals 15 volts divided by 10 ohms equals 1.5 amps.

Now that we have values for both 𝐼 one and 𝐼 two, we can calculate the value of 𝐼 using Kirchhoff’s first law. We find that 𝐼 equals 𝐼 one plus 𝐼 two equals three amps plus 1.5 amps equals 4.5 amps. This value corresponds with option (C), so the value of 𝐼 must equal 4.5 amps. Any other value of 𝐼 would violate the conservation of charge and the conservation of energy in a circuit. So the correct answer is option (C).

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