Question

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

Let the net resistance after adding the resistor be $R^{\prime }$.
Using Ohm's law:
$V=I{R}^{\prime }$
$R^{\prime }=\frac{6}{0.25}\Omega$
$R^{\prime }=24\Omega$

Note that $R^{\prime }>R$. This means the new resistor has to be added in series with the lamp. Let the resistance to be added be $R^{\prime \prime }$.
Equivalent of series connection of lamp and $R^{\prime \prime }$ will be $R^{\prime }$.
$R^{\prime }=R+R^{\prime \prime }$
$24=20+R^{\prime \prime }$
$R^{\prime \prime }=4\Omega$

Therefore, the extra resistance to be added is $4\Omega$.