Question

A current of 1 ampere flows in a series circuit having an electric lamp and a conductor of 5Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5Ω conductor and potential difference across the lamp will take place? Give reason.

Answer 1

Let the resistance of the electric lamp be R_lamp.

Current (I) = 1 A

Resistance of conductor (R_conductor)  = 5Ω

Potential difference of battery (V) = 10 V

Since the lamp and the conductor is connected in series, thus same current 1 A will pass through both of them.

Using Ohm's law,

R_net = \(\frac VI\)

R_net = \(\frac{10}1\)

R_net = 10Ω

R_net = R_lamp + R_conductor

⇒ 10 = R_lamp + 5

⇒ R_lamp = 5Ω

Potential difference across lamp,

V_lamp = I × R_lamp = 1 × 5 = 5V

When 10Ω resistor is connected parallel to the series combination of lamp and conductor (R_net = 5 + 5 = 10Ω) then the equivalent resistance,

\(\frac 1{R_{eq}} = \frac 1{10} + \frac 1{10} \)

\(= \frac 2{10}\)

\(= \frac 15\)

⇒ R_eq = 5Ω

Using Ohm's law,

\(I' = \frac V{R_{eq}}\)

⇒ \(I' = \frac{10}5\)

⇒ \(I' = 2A\)

Current will distribute equally in two parallel parts.

Thus,\(\frac{I'}2 = 1A\) current will pass through both the lamp and the resistor of 5Ω (because they are connected in series).

Potential difference across the lamp (R_lamp) = 5Ω

V'_lamp = 1 × 5 = 5V

Hence, there will be no change in current through the conductor of resistance 5Ω, and potential difference across the lamp.

Answer 2

(ii) Now, a resistance of 10Ω is connected in parallel with the series combination. Therefore, the total resistance of the circuit is given by

Thus, I A current will flow through 10Ω resistor and 1A will flow through the lamp and conductor of 5Ω resistance. Hence, there will be no change in current flowing through 5Ω conductor. Also, there will be no change in potential difference across the lamp.