Question Video: Calculating the Difference in Potential Difference across Circuit Components | Nagwa Question Video: Calculating the Difference in Potential Difference across Circuit Components | Nagwa

# Question Video: Calculating the Difference in Potential Difference across Circuit Components Physics • Third Year of Secondary School

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The circuit diagram represents a galvanometer combined with a shunt resistor. The emf of the cell connected to the galvanometer and the shunt is 3.0 V. The diagram does not represent a circuit in which the galvanometer and shunt function correctly as an ammeter. What is the difference between the potential difference across the shunt and the potential difference across the galvanometer?

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### Video Transcript

The circuit diagram represents a galvanometer combined with a shunt resistor. The emf of the cell connected to the galvanometer and the shunt is 3.0 volts. The diagram does not represent a circuit in which the galvanometer and shunt function correctly as an ammeter. What is the difference between the potential difference across the shunt and the potential difference across the galvanometer?

So, in this question weβre comparing potential differences across two components that are connected in parallel: a galvanometer and a shunt resistor. We can label the potential difference across the galvanometer π πΊ and the potential difference across the shunt resistor π π.

Now, the question tells us that the emf of the circuitβs cell is 3.0 volts. However, we donβt know the resistance of the shunt resistor or the resistance of the galvanometer, nor do we know anything about the current in the circuit. Actually, though, we donβt need to know those values, or any other numerical values, to solve this question. All we need to do is recall the fact that parallel branches in a circuit have the same potential difference across them. Because of this, we know that π πΊ must be equal to π π.

Now, the question is asking us to find the difference between the potential difference across the shunt resistor and the potential difference across the galvanometer. Since we know that these quantities are equal, then there must be no difference between them. We can show this algebraically by subtracting π π from both sides of this equation, which gives us π πΊ minus π π equals zero. In other words, the difference between π πΊ and π π is zero.

Now, since we measure potential differences in volts, then our final answer for the difference between the potential difference across the galvanometer and the potential difference across the shunt resistor is zero volts.

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