Question

The figure shows a circuit
When the circuit is switched on, the ammeter reads $0.5A$.
(i) Calculate the value of the unknown resistor $R$.
(ii) Calculate the charge passing through the $3\Omega$ resistor in $120s$.
(iii) Calculate the power dissipated in the $3\Omega$ resistor.

Answer 1

Given:

Ammeter reading = Current (I) = $0.5A$

Voltage (V) = $6.0volt$

(i) $\text{Calculating the unknown resistance (R):}$

In the circuit, two resistors $R$ and $3\Omega$ are in series. Thus the equivalent resistance is $R_{eq}=R+3$

We know,

Electric current $\left(I\right)=\frac{Totalpotentialdifference\left(V\right)}{Totalresistance\left(R_{eq}\right)}$
$\Rightarrow 0.5=\frac{6}{3+R}$
$\Rightarrow R+3=\frac{60}{5}$
$\Rightarrow R=12-3=9\Omega$

The unknown resistor, $R=9\Omega$

(ii) $\text{Calculating the charge (q) passing through the 3 ohm resistor:}$

We know from the definition of current, $i=\frac{q}{t}$

$\therefore q=i\times t=0.5\times 120=60C$
Charge passing through the resistor, $q=60C$

(iii) $\text{Calculating the power dissipation in the 3 ohm resistor:}$

We know from the definition of power, $P=V\times I$ $=I^{2}R$

$\therefore P={\left(0.5\right)}^2\times 3=0.25\times 3$

The power dissipated, $P=0.75W$