A student kept his 9.0 V, 7.0 Wradio turned on at full volume from 9:00 P.Muntil 2:00 A.M.. How much charge went through it?

1. The potential difference applied, V = 9.0V
2. The power of the radio P = 7.0W
3. The time for which radio is kept on t = 5.0hr(9.00 p.m. to 2.00 a.m.)

The power or rate of energy transfer, in an electrical device across which a potential difference is equal to the product of current and potential difference. It can also be defined as energy transferred per unit time.

Current through the radio can be calculated from its power (energy dissipation rate) and the potential difference across it using the corresponding relation. The amount of charge that went through the radio can be calculated from the current and the time for which the radio is kept on.

Formulae:

The electric power in the circuit of a potential difference, P = IV …(i)

Here, P is the power, l is the current, V is the potential difference.

The current flowing through the circuit,$l=\frac{q}{t}$ …(ii)

Here, l is current, q is the charge, and t is the time.

Using the given data in equation (i), we can get the amount of current flowing into it as follows:

Now, using this above value of current, we can get the amount of charge that went through it by using equation (ii) as follows:

($Here,t=5.0hror1.8\times {10}^{4}s$)

$\begin{array}{l}q=\left(\frac{7.0}{9.0}A\right)\times \left(1.8\times {10}^{4}s\right)\\ =\frac{1.26\times {10}^5}{89.0}\\ =1.4\times {10}^{4}C\end{array}$

Hence, the value of the charge is $1.4\times {10}^{4}C$.