Question

A current $500mA$flows in a series circuit containing an electric lamp and a conductor of $10\Omega$ when connected to $6V$ battery. Find the resistance of the electric lamp.

Answer 1

Step 1: Given data

Electric current flowing through the circuit, $i=500mA=0.5A$

Resistance of the conductor, $R_c=10\Omega$

Potential difference across the battery, $V=6v$

Resistance of the lamp, $R_l=?$

Equivalent resistance of the given series circuit, $R_s=?$

Step 2: Calculation of the resistance of the lamp

The circuit diagram of the given circuit is shown below

Since the conductor and the bulb are connected in series, therefore the total resistance in this series circuit will be equal to the sum of the resistance of the lamp and the conductor.

$R_s=R_l+R_c$ ……………………..(a)

Substituting the given value of the resistance of the resistor in equation (a), we get

$R_s=R_l+10$ ……………………….(b)

Applying ohm's law to the given circuit, we get

$V=i{R}_s$ ……………………………(c)

Using equation (b) and the given value of electric current in equation (c), we get

$6v=0.5A\times (R_l+10)$

$$\begin{gathered} 12=R_l+10 \\ {R}_l=2\Omega \end{gathered}$$

Hence, the resistance of the electric lamp is equal to $2\Omega$