Question

Current in a simple series circuit is $5A$. When an additional resistance of 2$\Omega$ is inserted, the current drops to $4A$. What was the resistance of the original circuit ?

• A. 8 $\Omega$
• B. 6 $\Omega$
• C. 10 $\Omega$
• D. 4 $\Omega$

Answer 1

Answer: (A)

Let V be the applied potential difference and R be the resistance of original circuit.
Then $\frac{V}{R}$ = I = 5A ...$(i)$
with the added resistance of 2$\Omega$, total resistance becomes (R + 2)$\Omega$
Then $\frac{V}{(R+2)}$ = I' = 4 A ...$(ii)$
Dividing equation $(i)$ by equation $(ii)$ $\frac{R+2}{R}=\frac{5}{4}$ or 4R + 8 = 5R
or $Originalresistance,$ R = 8 $\Omega$