Question

A battery of emf 10V and internal resistance \[3\,\Omega \] is connected to a resistor. The current in the circuit is 0.5A. The terminal voltage of the battery when the circuit is closed is
A. 10 V
B. 0 V
C. 8.5 V
D. 1.5 V

Answer 1

Hint: Using Kirchhoff’s voltage, express the current in the circuit and therefore, determine the value of resistance. Using Ohm’s law, express the voltage across the resistance. The voltage across the resistance is the terminal voltage of the battery.

Formula used:
Ohm’s law, \[V = IR\],
where, V is the voltage, I is the current and R is the resistance.

Complete step by step answer:
We have given that a supply of emf 10V is given to the circuit containing a resistor of \[3\,\Omega \]. The current in the circuit is 0.5A. The above mentioned circuit will look like as shown in the figure below.

Let the internal resistance of the battery is r. Applying Kirchhoff’s voltage law in the closed loop, we get,
\[E - Ir - IR = 0\]
\[ \Rightarrow E = I\left( {r + R} \right)\]
\[ \Rightarrow I = \dfrac{E}{{r + R}}\]
This is the magnitude of the current that will flow through the circuit.
Rearranging the above equation for r, we get,
\[r + R = \dfrac{E}{I}\]
\[ \Rightarrow R = \dfrac{E}{I} - r\]
Substituting 10V for E, 0.5A for I and \[3\,\Omega \] for r in the above equation, we get,
\[R = \dfrac{{10}}{{0.5}} - 3\]
\[ \Rightarrow R = 17\,\Omega \]
Now, we can express the terminal voltage in the circuit as voltage drop across the resistance R.
\[V = IR\]
Substituting 0.5A for I and \[17\,\Omega \] for R in the above equation, we get,
\[V = \left( {0.5} \right)\left( {17} \right)\]
\[ \therefore V = 8.5\,{\text{V}}\]
Thus, the terminal voltage of the battery when the circuit is closed is equal to 8.5 V.

So, the correct answer is option C.

Note: While using Kirchhoff’s law, if there is drop in the voltage across the component of the circuit, the voltage should be taken as negative and if there is addition of the voltage. The internal resistance of the battery is negligible and therefore, we often neglect its value. The direction of the conventional current in the circuit is always from the positive terminal of the battery towards its negative terminal.