Question

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the
circuit is closed?​

Answer 1

I=E/r+R here I is 0.5A , E is 10v , r=3 ohm and R=? then , 0.5=10/3+R, 1.5+0.5R=10,. 8.5=0.5R, R=8.5/0.5, R=17 ohm , terminal voltage = I×R, terminal voltage (V)= 0.5×17 , V=8.5 volt

Answer 2

⚡Answer :- ⚡

V = 8.5v

⚡Explanation :- ⚡

Given:

$\implies$ The EMF of the battery (E = 10 V)

$\implies$ The internal resistance of the battery (R = 3 Ω)

$\implies$ The current in the circuit (I = 0.5 A)

______________________

Consider the resistance of the resistor to be R.

The current in the circuit can be found out using Ohm’s Law as,

$\rm \: \red{ \underline{ \boxed{I \: = \rm \: \frac{E}{R + r} }}}$

Consider the Terminal voltage of the resistor to be V.

Then, according to Ohm’s law,

${\red{ \rm{ \underline{ \boxed{ \rm \: V = IR}}}}}$

Substituting values in the equation, we get

$\implies$ V = 0.5 × 17

$\implies$ V = 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is 8.5 V.