Now if the resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in the current flowing through the 5 Ω conductor and potential difference across the lamp will take place? Give reason.
Let R1 be the resistance of the electric lamp and R2 the resistance of a conductor.
In series, the total resistance R = R1 + R2 = 5 + R1
Current (I) = 1 A
Voltage (V) = 10 V
Using Ohm’s law
I = V/R
1 = 10/(5+R)
5 + R = 10
R = 5Ω
Now, a 10 Ω resistance is connected in parallel with the series combination. Therefore, the total resistance of the circuit is calculated as –
1/Rp = 1/(R1 + 5) + 1/10
Rp = 5Ω
Therefore, the current flowing in the circuit is equal to I = V/R
I = 10/5
I = 2A
Read More: Ohm’s Law and its Limitations
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