Question

An electric lamp of resistance 20Ω and a conductor of resistance 4Ω are connected to a 6V battery as shown in the circuit. Calculate:
(a) The total resistance of the circuit
(b) The current through the circuit
(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and
(d) Power of the lamp.

Answer 1

Step 1: Given data

1. The resistance of the electric lamp is, $R_L=20\Omega$.
2. The resistance of the conductor is, $R_c=4\Omega$.
3. Supplied voltage, $V=6volt.$

Ohm's law:

1. Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across the conductor.
2. The mathematical form of Ohm's law is defined by the form, $V=IR$, where, V is the potential difference, I is the current flowing through the circuit and R is the resistance of the conductor.

Power:

1. Power is the amount of energy consumed by a circuit in a unit time.
2. Power is defined by the form, $P=VI$, where, V is the voltage across a circuit and I is the current flow through the circuit.

Diagram:

Step 2: (a) Calculation of the resistance

From the figure, the conductor and the electric lamp are in a series combination. So the resistance of the coil is

$$\begin{gathered} R_T=R_L+R_C=20+4 \\ or{R}_T=24\Omega \end{gathered}$$

Therefore, the resistance of the coil is $24\Omega$.

Step 3: (b) Calculation of the circuit current

As we know, $V=IR$

So,

$$\begin{gathered} I=\frac{V}{R}=\frac{6}{24}=0.25 \\ orI=0.25A \end{gathered}$$

Therefore, the current through the circuit is $0.25A.$

Step 4: (c) Calculation of the lamp and conductor voltage

Again we know, $V=IR$

So, the potential difference across the conductor is

$$\begin{gathered} V_c=0.25\times 4=1 \\ or{V}_c=1volt. \end{gathered}$$

And the potential difference across the lamp is

$$\begin{gathered} V_L=0.25\times 20=5 \\ or{V}_L=5volt. \end{gathered}$$

Step 5:(d) Calculation of lamp power

We know, that power of a circuit is $P=VI$

So,

$$\begin{gathered} P=5\times 0.25=1.25 \\ orP=1.25watt. \end{gathered}$$ (since the voltage across the lamp is $V_L=5volt$ and $0.25A$ is the circuit current.)

So, the power of the lamp is $1.25watt$.