Question

In the figure, the value of resistor R is $(25+\frac{1}{2})\Omega$, where I is the current in amperes. The current I is___.

• A. 10

Answer 1

The correct option is A 10
Given,
$R=(25+\frac{1}{2})\Omega$
or,
$I=(2R-50)$


Applying KVL is given loop,
we have:
$300=IRor300=(2R-50)\times R$

we get,
$R=30\Omega or-5\Omega$

Since resistance can't be negative. Therefore,

$R=30\Omega$

$Hence,I=(2R-50)$
$=(2\times 30-50)A$
$=10A$

$\therefore Current,I=10A$