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Question Video: Using Kirchhoff’s Second Law to Calculate Voltage in a Parallel Circuit Physics

The resistor in the circuit shown is powered by two batteries connected in parallel. One battery has a terminal voltage of 2.5 V. What must be the terminal voltage of the other battery for the potential drop across the resistor to be determinable?

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Video Transcript

The resistor in the circuit shown is powered by two batteries connected in parallel. One battery has a terminal voltage of 2.5 volts. What must be the terminal voltage of the other battery for the potential drop across the resistor to be determinable?

To answer this question, we will use Kirchhoff’s second law. Recall that Kirchhoff’s second law states that the sum of the potential differences across each component in a loop in a circuit is equal to zero. Consider the loop that we’ve highlighted in the part of the circuit containing the batteries in parallel. By applying Kirchhoff’s second law to this loop, we find that 2.5 volts minus the unknown voltage 𝑉 of the battery on the lower branch must equal zero.

We can solve this equation for the unknown voltage by adding 𝑉 to both sides. We then have that 2.5 volts is equal to the unknown voltage 𝑉. That is, the terminal voltage of the battery on the lower branch is equal to 2.5 volts. This makes sense because in order to obey Kirchhoff’s second law, batteries in parallel need to have identical terminal voltage. Thus, because one of the batteries in parallel has a terminal voltage of 2.5 volts, we know that the terminal voltage of the other battery must also be 2.5 volts.

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