Question

An ideal voltmeter V is connected to a 2.0 - $\mathbf{\Omega }$resistor and a battery with emf 5.0 V and internal resistance 0.5$\mathbf{\Omega }$as shown in Figure. (a) What is the current in the resistor 2.0 -$\mathbf{\Omega }$? (b) What is the terminal voltage of the battery? (c) What is the reading on the voltmeter? Explain your answers.

Short answer

1. The current in the resistor is 0 A
2. The terminal voltage of the battery is 5.0 V
3. The reading on the voltmeter is 5.0 V

Step-by-step solution

Determine the current in the external circuit

(a) As shown in the given figure, the voltmeter is connected in the path of the resistance R and the battery, so it is part of the path. This voltmeter is an ideal voltmeter, so its internal resistance is infinite, which means there will be a large resistance which will prevent the current to flow through the circuit.

Thus, the current in the resistance $2.0\mathrm{\Omega }$is zero, I = 0

Determine the terminal voltage of the battery

(b) As there is no current flowing through the circuit, we get that the terminal voltage equals emf of the battery as the current is zero.

Thus, the terminal voltage of the battery is 5.0 V.

Determine the voltmeter reading

(c) As there is no voltage drop and the voltage across the circuit is same for the battery, therefore the voltmeter will read the same value for the emf of the battery.

Therefore, the voltmeter reading is equal to the emf of the battery which means that the reading shows 5.0 V .