Question

The current in a simple series circuit is $5.0\, amp$. When an additional resistance of $2.0\, ohms$ is inserted, the current drops to $4.0\, amp$. The original resistance of the circuit in ohms was

• A. 1.25
• B. 8
• C. 10
• D. 20

Answer 1

Answer: (B)

Let the original resistance is $R \Omega$.
$V =I R $
$V =5 \times R $
$=5 R \,\,\,\,\,\dots(i)$
When $2 \Omega$ resistance is inserted, then total resistance $=(R+2) \Omega$
$V =I'(R+2) $
$=4(R+2)\,\,\,\,\,\dots(ii)$
From Eqs. (i) and (ii), we get
$5 R=4(R+2)$
$R=8 \,\Omega$