A charge of 24 C passes through a circuit every 9 s. If the circuit can generate 6 W of power, what is the circuit's resistance?

May 30, 2017

Resistance is defined as the ability of the conductor to obstruct the flow of electric current through it. Resistance is basically defined by Ohm's Law as:-
$V = I R$

Explanation:

In the given question, we have:-
Charge = $24 C = Q$
Time = $9 s = t$

Rate of flow of current thorugh a cross-section of a conductor is givem by the following relation:-

$Q = I t$ where,
Q = Given charge
I = Current that flows
t = time (in second)

So, putting above stated values in this relation, we get:-

$24 = I . 9$
or $I = \frac{24}{9} A$

Power is defined as the ability to do work. Power is given by the formula:-

$P = V I$ where,
P = Power in watt
V = Potential difference across the circuit
I = Current through the circuit

Putting values in the relation, we get:-

$6 = V . \frac{24}{9}$
or $V = 6 \cdot \frac{9}{24}$ V

As stated above, the expression for Ohm's Law gives resistance as:-
$R = \frac{V}{I}$

Hence, using the values obtained for V and I , we get:-

$R = 6 X \frac{9}{24} \cdot \frac{9}{24}$
or  R = 0.84 Ω

Hence, the resistance of the circuit is 0.84 Ω.

May 30, 2017

The answer is $= 843.75 m \Omega$

Explanation:

We apply the equation

$Q = I t$

The charge is $Q = 24 C$

The time is $t = 9 s$

The current is $I = \frac{Q}{t} = \frac{24}{9} = \frac{8}{3} A$

The power

$P = U I$

But according to Ohm's Law

$U = R I$

So,

$P = R I \cdot I = R {I}^{2}$

So,

the resistance is $R = \frac{P}{I} ^ 2$

$= \frac{6}{\frac{8}{3}} ^ 2 = 0.84375 \Omega$