A charge of $40 C$ $40 C$ passes through a circuit every $8 s$ $8 s$ . If the circuit can generate $16 W$ $16 W$ of power, what is the circuit's resistance?

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Answer 1 · 2016-09-11T13:32:19

$R = 0.64$ $R = 0.64$ Ω

One of the formulas for electric power is $P = I^{2} R$ $P = I^(2) R$ .

Also, electric current is defined as the flow of charge.

Mathematically, this is expressed as $I = \frac{Q}{t}$ $I = (Q) / (t)$ .

Let's substitute this expression of $I$ $I$ into the equation for electric power:

$\Rightarrow P = {(\frac{Q}{t})}^{2} R$ $=> P = ((Q) / (t))^(2) R$

$\Rightarrow P = \frac{Q^{2}}{t^{2}} R$ $=> P = (Q^(2)) / (t^(2)) R$

Then, let's rearrange the equation to express it in terms of $R$ $R$ :

$\Rightarrow R = (\frac{P t^{2}}{Q^{2}})$ $=> R = ((P t^(2)) / (Q^(2)))$ Ω

Now, let's substitute the values for electric charge, time and electric power:

$\Rightarrow R = (\frac{16 \cdot 8^{2}}{40^{2}})$ $=> R = ((16 cdot 8^(2)) / (40^(2)))$ Ω

$\Rightarrow R = (\frac{16 \cdot 64}{1600})$ $=> R = ((16 cdot 64) / (1600))$ Ω

$\Rightarrow R = (\frac{1024}{1600})$ $=> R = ((1024) / (1600))$ Ω

$\Rightarrow R = 0.64$ $=> R = 0.64$ Ω

A charge of 40C40 C passes through a circuit every 8s8 s. If the circuit can generate 16W16 W of power, what is the circuit's resistance?