Question

The current in a simple series circuit is $5.0amp$. When an additional resistance of $2.0ohms$ is inserted, the current drops to $4.0amp$. 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 $R\Omega$.
$V=IR$
$V=5\times R$
$=5R\dots (i)$
When $2\Omega$ resistance is inserted, then total resistance $=(R+2)\Omega$
$V=I^{\prime }(R+2)$
$=4(R+2)\dots (ii)$
From Eqs. (i) and (ii), we get
$5R=4(R+2)$
$R=8\Omega$