A battery of e.m.f. 10V and internal resistance of 3 $\Omega$ is connected to a resistor. If the current in the circuit is 0.5A. What is the resistance of the resistor?

Question

Accepted Answer

The correct option is (3):

To find the resistance of the external resistor, we need to apply the concept of Ohm's Law for a circuit containing a battery with internal resistance. The total electromotive force (e.m.f.) of the battery, denoted by $E$, drives the current through both the internal resistance of the battery ($r$) and the external resistance ($R$). The relationship between these quantities is given by the formula:
$E = I (R + r)$
where $I$ is the current flowing through the circuit.

From the question, we are given:
E.m.f. of the battery, $E = 10 	ext{ V}$
Internal resistance of the battery, $r = 3 	ext{ } \Omega$
Current in the circuit, $I = 0.5 	ext{ A}$

We need to find the external resistance, $R$.
Rearranging the formula to solve for $R$:
$R + r = \frac{E}{I}$
$R = \frac{E}{I} - r$

Now, substitute the given values into the equation:
$R = \frac{10 	ext{ V}}{0.5 	ext{ A}} - 3 	ext{ } \Omega$
$R = 20 	ext{ } \Omega - 3 	ext{ } \Omega$
$R = 17 	ext{ } \Omega$

Therefore, the resistance of the external resistor is $17 	ext{ } \Omega$. This calculation correctly accounts for both the e.m.f. and the internal resistance of the battery in determining the total resistance in the circuit.

The incorrect option is (1): $15 	ext{ } \Omega$. This would be the result if one incorrectly subtracted the internal resistance from the e.m.f. value before dividing by the current, or made a computational error in the final step.
The incorrect option is (2): $14 	ext{ } \Omega$. This value is likely the result of an arithmetic error in the calculation.
The incorrect option is (4): $20 	ext{ } \Omega$. This would be the total resistance of the circuit ($R\_{total} = E/I$) without accounting for the internal resistance, or an incorrect assumption that the internal resistance is zero.