A bulb is joined to a battery of emf $6\,V$. A steady current of $0.5\,A$ flows through the circuit. Calculate the cost of electricity at Rs. \[9.50/-\] per \[kWh\] for 5 minutes.

Question

A bulb is joined to a battery of emf $6\,V$. A steady current of $0.5\,A$ flows through the circuit. Calculate the cost of electricity at Rs. $$9.50/-$$ per $$kWh$$ for 5 minutes.

Vedantu Content Team · Accepted Answer

Hint: Learn the Joule’s law of heating for resistive circuits and find the energy loss by the bulb to find the cost of electricity. The Joule’s heating for a resistive circuit is proportional to the square of current flowing, the resistance of the circuit and the time. Formula used: The Joule’s law of heating in an electric circuit is given by, $$H = {I^2}Rt$$ where, $$I$$ is the current through the circuit, $$R$$ is the resistance of the circuit and $$t$$ is the time passed.  Complete step by step answer: We have given here a bulb which is joined to a battery of emf 6V and the current through the circuit is 0.5A. It glows for 5 minutes constantly. Now, we have to find the electricity cost at a rate of Rs. 9.50. We know that the loss of electric energy in a circuit is nothing but the Joule’s heating from the circuit. Now, Joule’s law of heating in an electric circuit is given by, $$H = {I^2}Rt$$ Hence, first we have to find the resistance of the bulb. We have given, $$I = 0.5A$$$$V = 6V$$.So, from Ohm’s law resistance $$R$$is equal to, $$R = \dfrac{V}{I} = \dfrac{6}{{0.5}} = 12\Omega $$So, putting the values of $$I = 0.5A$$, $$R = 12\Omega $$ and $$t = 5 \times 60 = 300\,s$$We have from joule’s law, $$H = {(0.5)^2} \times 12 \times 300$$$$\Rightarrow H = 900$$Hence, the energy loss due to the bulb is,$$900\,J = 900\,Ws = \dfrac{{900}}{{3600 \times 1000}}\,kWh$$$$\Rightarrow 900\,J= 0.0025kWh$$Hence, the cost of electricity will be, $$Rs.\,0.0025 \times 9.50 = Rs.0.023 = 2.375\,paise$$Hence, the cost of electricity is $$2.375\,paise$$.Note: When converting the energy from$$J$$ to in terms of $$kWh$$make sure to multiply the correct constants with the denominator else the result will be flawed. The expression of the Joule’s heating is the same as the power generation of a circuit,$$P = VI$$.We can rewrite this using the Ohm’s law $$V = IR$$as $$P = {I^2}R$$. The only difference is that in the expression of power the expression of time is not there.