Question

A 6 V battery is connected across a lamp whose resistance is 20 $\Omega$. Calculate the value of the extra resistance in $\Omega$ which must be used in the circuit to get 0.25 A current in the circuit.

• A. 4 $\Omega$
• B. 6 $\Omega$
• C. 2 $\Omega$
• D. 8 $\Omega$

Answer 1

The correct option is A

4 $\Omega$

Given:
Lamp resistance, R = 20 $\Omega$
Applied voltage, V = 6V
Desired current in the circuit, I = 0.25A
Extra resistance to be added = R'
Now,
Total resistance in the circuit, $R_s=R+R^{\prime }$ $\Omega$.
From Ohm's law, $R_s=\frac{V}{I}$
Putting the values in the equation, $R_s=\frac{6}{0.25}$
$=24\Omega$
Now, $R_s=R+R^{\prime }$ $\Omega$
So, $R^{\prime }=R_s-R$
$R^{\prime }=24\Omega$ - 20$\Omega$
$R^{\prime }=4\Omega$
Therefore extra resistance to be added is = 4$\Omega$.