Question

A nonideal 12 V battery is connected in a circuit with a single resistor of resistance 6 Ω. A voltmeter connected across the resistor reads 9 V. What is the internal resistance of the battery?

a. 2 Ω
b. 3 Ω
c. 4 Ω
d. 6 Ω
e. 14 Ω

Answer 1

The internal resistance of the nonideal 12 V battery connected with a 6 Ω resistor is calculated to be 2 Ω, taking into account the measured voltage across the resistor and the provided emf of the battery. The correct answer is option a.

Step-by-step explanation:

The question pertains to the calculation of the internal resistance of a nonideal battery when it is connected to a circuit with a resistor. The situation describes a 12 V nonideal battery connected in series with a 6 Ω resistor, and the voltmeter reading across the resistor is 9 V. To find the internal resistance of the battery, we can use Ohm's Law and the definition of terminal voltage.

First, we calculate the current (I) flowing through the circuit using the voltage across the resistor (V) and its resistance (R):

I = V / R

I = 9 V / 6 Ω

I = 1.5 A

Since the voltage across the resistor is less than the battery's emf, the internal resistance is causing a voltage drop. We can find this voltage drop (Vdrop) across the internal resistance:

Vdrop = emf - V

Vdrop = 12 V - 9 V

Vdrop = 3 V

Finally, we calculate the internal resistance (r) of the battery using the current (I) and the voltage drop (Vdrop):

r = Vdrop / I

r = 3 V / 1.5 A

r = 2 Ω

Therefore, the internal resistance of the battery is 2 Ω, which corresponds to option (a).