An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate:

(a) the total resistance of the circuit

Given:

Resistance of electric lamp, $R_1=20\Omega$

Resistance of conduxtir, $R_2=4\Omega$

Voltage, $V=6V$

To find: 

(a) the total resistance of the circuit, $R_T$.

(b) the current through the circuit,

(c) the potential difference across the (i) electric lamp and (ii) conductor, and

(d) power of the lamp.

Solution: (a)

Here, the resistors, i.e lamp and conductor are connected in series.

We know that equivalent or total resistance in series is given as-

${R_T}={R_1}+{R_2}$

Substituting the value of ${R_1}\ and\ {R_2}$ in the formula we get-

${R_T}=20+4$

${R_T}=24\Omega$

Thus, the total resistance of the circuit is 24 Ohm.

Solution: (b)

From Ohm's law, we know that-

$V=I\times R$

It can be rearranged for $I$ as-

$I=\frac {V}{R}$

Putting the value of $V\ and\ R$ we get-

$I=\frac {6}{24}$              $(\because R=R_{T})$

$I=\frac {1}{4}$  

$I=0.25A$  

Thus, the current through the circuit is 0.25 Ampere.

Solution: (c) (i) The potential difference across the electric lamp.

From Ohm's law, potential difference, $V$ is given as-

$V=I\times R$

Putting the value of $I=0.25$ and $R=20$ we get-

$V=0.25\times 20$

$V=5V$

 (i) The potential difference across the conductor.

From Ohm's law, potential difference, $V$ is given as-

$V=I\times R$

Putting the value of $I=0.25$ and $R=4$ we get-

$V=0.25\times 4$

$V=1V$

Solution: (d) 

Formula of power is given as-

$P=I^2\times R$

Putting the value of $I=0.25$ and $R=20$ we get-

$P=(0.25)^2\times 20$

$P=0.0625\times 20$

$P=1.25W$

Thus, the power of the lamp is 1.25 W.