Electric Circuits and Resistance: Question 2

Worked solution

Part (a): Combined resistance

In a series circuit, resistances simply add together:

$R_{\text{total}} = R_1 + R_2$

$R_{\text{total}} = 8\,\Omega + 4\,\Omega = \boxed{12\,\Omega}$

Part (b): Current shown on the ammeter

Rearrange $R = V/I$ to make current the subject:

$R = \frac{V}{I} \quad\Rightarrow\quad I = \frac{V}{R}$

The battery’s e.m.f. drives the current around the whole circuit, so use the total resistance found in part (a):

$I = \frac{6.0\text{ V}}{12\,\Omega}$

$I = \boxed{0.50\text{ A}}$

Because this is a series circuit, this same current of $0.50\text{ A}$ flows through both $R_1$ and $R_2$.

Part (c): Potential difference across $R_2$

Use $V = IR$ with the current from part (b) and the resistance of $R_2$ only:

$V_2 = I \times R_2 = 0.50\text{ A} \times 4\,\Omega$

$V_2 = \boxed{2.0\text{ V}}$

(As a check: the p.d. across $R_1$ would be $0.50 \times 8 = 4.0\text{ V}$, and $4.0 + 2.0 = 6.0\text{ V}$, which matches the e.m.f. of the battery, as it must, since the p.d.s across resistors in series add up to the source p.d.)

Final answers

• (a) Combined resistance $=$ $12\,\Omega$
• (b) Ammeter reading $=$ $0.50\text{ A}$
• (c) P.d. across $R_2$ $=$ $2.0\text{ V}$