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NEET PHYSICSEasy

A battery of emf 10 V and internal resistance 3 \Omega is connected to a resistor. If the current in the circuit is 0.5 A, what is the terminal voltage of the battery when the circuit is closed?

A

10 V

B

8.5 V

C

1.5 V

D

7.2 V

Step-by-Step Solution

According to the NCERT text (Section 3.10), the terminal voltage (VV) across a cell when current (II) is drawn from it is given by the formula V=εIrV = \varepsilon - Ir, where ε\varepsilon is the EMF and rr is the internal resistance. Substituting the given values: V=10 V(0.5 A×3 Ω)=101.5=8.5 VV = 10 \text{ V} - (0.5 \text{ A} \times 3 \ \Omega) = 10 - 1.5 = 8.5 \text{ V}. This question corresponds directly to Exercise 3.2 in the textbook , .

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