A current of 5 A flows through an electrical appliance when the potential difference of 50 V is applied across its terminals What will be the current drawn by the appliance if the potential difference is increased to 110 V?

Correct option is D

Sol:  The correct answer is (d) 11A

Given:

Initial current (I₁) = 5 A

Initial potential difference (V₁) = 50V

New potential difference (V₂) = 110V

Formula Used:

Ohm's Law V = I $$\times$$ R

$V$ = Potential difference (Voltage) in volts

$I$ = Current in amperes

$R$ = Resistance in ohms

Solution:

Using the initial values of current and voltage, calculate the resistance of the appliance-

R = $$\frac{V_1}{V_2}$$ = $$\frac{50}{5}$$ 10 Ω

​New Current (I₂)

new potential difference =110V 

​$$I_2$$ = $$\frac{V_2}{R}$$ = $$\frac{110}{10}$$ =11A

​The new current drawn by the appliance when the potential difference is increased to 110 V is 11 A.