Question Video: Using Kirchhoff’s Laws to Analyze Combination Circuits | Nagwa Question Video: Using Kirchhoff’s Laws to Analyze Combination Circuits | Nagwa

Question Video: Using Kirchhoff’s Laws to Analyze Combination Circuits Physics • Third Year of Secondary School

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What is the current 𝐼 in the circuit shown? Give your answer to one decimal place.

06:11

Video Transcript

What is the current 𝐼 in the circuit shown? Give your answer to one decimal place.

Looking at the circuit diagram, we can see that there are two loops, each containing a cell and a resistor. Because of the way this circuit has been put together, the resistors are neither in series nor in parallel with each other. This means it’s not possible to work out the total resistance or equivalent resistance of the 16-ohm resistor and the 11-ohm resistor combined. Instead, we can tackle this question using Kirchhoff’s second law, which is also known as Kirchhoff’s loop rule.

We can begin by labeling each of the components in the circuit. Let’s call the resistance of the 16-ohm resistor 𝑅 one and the resistance of the 11-ohm resistor 𝑅 two. We’ll also say the potential difference of the cell on the left is 𝑉 one and the potential difference of the cell on the right is 𝑉 two. Next, we label the currents in each branch of the circuit.

Here, the orientation of the cells tells us which way the conventional current goes. On the left, the positive terminal, denoted by the longer line, is on top of the cell and the negative terminal, denoted by the shorter line, is on the bottom. This means that the direction of conventional current in the left loop goes this way. Let’s label this current 𝐼 one. Likewise, in the loop on the right, we have conventional current going in this direction. Let’s call this current 𝐼 two. And remember, in the middle, we have the current 𝐼 pointing downward.

Now that we’ve labeled all of components and currents in our circuit diagram, we can recall that Kirchhoff’s second law states that the sum of the potential differences across all of the components in a loop is always equal to zero. Let’s first apply this to the loop on the left. We’ll call this loop one. Applying Kirchhoff’s second law to this loop, we can say that the sum of the potential differences across this cell and this resistor must be zero.

We can label the potential difference across this resistor 𝑉 𝑅 one. Then Kirchhoff’s second law tells us that 𝑉 one minus 𝑉 𝑅 one must be equal to zero. Notice that the potential difference provided by the cell, that’s 𝑉 one, is positive, while the potential difference across the resistor, 𝑉 𝑅 one, is negative. This is because there is a decrease in potential across the resistor. Now we can rearrange this equation slightly to give us 𝑉 𝑅 one equals 𝑉 one.

Now we know that the potential difference provided by this cell, 𝑉 one, is 22 volts. Therefore, thanks to Kirchhoff’s second law, we can see that 𝑉 𝑅 one must also be equal to 22 volts. Once we know this, we can use Ohm’s law to calculate the current 𝐼 one through the 16-ohm resistor. First, recall that Ohm’s law can be written as 𝑉 equals 𝐼𝑅, where 𝑉 is the potential difference across a resistor. 𝐼 is the current through the resistor. And 𝑅 is the resistance of that resistor.

Since we want to calculate the current through our resistor, we need to rearrange this to make 𝐼 the subject. To do this, we divide both sides of the equation by 𝑅 to give us 𝑉 over 𝑅 equals 𝐼. Then we can switch the left- and right-hand sides around to give us 𝐼 equals 𝑉 over 𝑅. Now in this specific case, we have 𝐼 one, the current through the 16-ohm resistor, is equal to 𝑉 𝑅 one, the potential difference across it, divided by 𝑅 one, its resistance. We’ve just worked out that the potential difference across the resistor, 𝑉 𝑅 one, is 22 volts. And we’re told in the question the resistance 𝑅 one is 16 ohms. Evaluating this expression, we get a result of 1.375 amps.

Let’s make a note of this over here. And we’re now ready to consider loop two. Now applying the same method as before, we can label the potential difference across the 11-ohm resistor as 𝑉 𝑅 two. Then by applying Kirchhoff’s second law to loop two, we find that 𝑉 two minus 𝑉 𝑅 two equals zero, which we can rearrange to give us 𝑉 𝑅 two equals 𝑉 two. Now the cell in loop two also provides a potential difference of 22 volts. Therefore, the potential difference across resistor 𝑅 two is also 22 volts.

Once again, we can now use Ohm’s law to calculate the current 𝐼 two, which passes through resistor 𝑅 two. So applying Ohm’s law to the resistor in loop two, we have 𝐼 two equals 𝑉 𝑅 two divided by 𝑅 two. We’ve worked out that 𝑉 𝑅 two is 22 volts. And we’re given that 𝑅 two is 11 ohms. 22 volts divided by 11 ohms is two amps. Let’s label this current on our diagram as well.

Okay, now that we have values for the currents 𝐼 one and 𝐼 two, we can use Kirchhoff’s first law to find the current 𝐼. We can recall that Kirchhoff’s first law, also known as Kirchhoff’s junction rule, states that the sum of the currents going into a junction or node in a circuit must be the same as the sum of the currents coming out of that junction.

Let’s apply this law to this junction in our circuit. We can see that we have two currents going into this junction: 𝐼 one and 𝐼 two. We also have just one current coming out: the current 𝐼, which we want to find. Kirchhoff’s first law says that the sum of the currents going into the junction, in this case that’s 𝐼 one plus 𝐼 two, is equal to the sum of the currents coming out of the junction, which in this case is just 𝐼. Substituting in our calculated values for 𝐼 one and 𝐼 two, we find that 𝐼 is equal to 1.375 amps plus two amps, which is equal to 3.375 amps.

The final thing we need to do to complete this question is to give the answer to one decimal place. And 3.375 to one decimal place is 3.4. So our final answer is that the current 𝐼 in the circuit shown is 3.4 amps.

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