D.C. Circuits: Question 10

Worked solution

Step 1: Combine the parallel resistors

For the $6.0\ \Omega$ and $3.0\ \Omega$ resistors in parallel: $\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0} = \frac{1}{2.0}$

So $R_p = 2.0\ \Omega$.

Check using decimals: $\frac{1}{6.0}=0.1667$ and $\frac{1}{3.0}=0.3333$; adding gives $0.1667+0.3333=0.500=\frac{1}{2.0}$, confirming $R_p=2.0\ \Omega$.

Step 2: Find the total resistance and current

The $3.0\ \Omega$ series resistor adds directly to $R_p$: $R_{\text{total}} = 3.0 + 2.0 = 5.0\ \Omega$

Since the internal resistance is negligible, the full e.m.f. drives the total current: $I = \frac{V}{R_{\text{total}}} = \frac{15}{5.0} = 3.0\text{ A}$

Check: $3.0\times5.0=15\text{ V}$, matching the battery’s e.m.f. Confirmed.

Step 3: Find the p.d. across the parallel section

$V_p = I\times R_p = 3.0\times2.0 = 6.0\text{ V}$

Check using the series resistor instead: the p.d. across the $3.0\ \Omega$ series resistor is $I\times3.0=3.0\times3.0=9.0\text{ V}$, and $9.0+6.0=15\text{ V}$, matching the e.m.f. Confirmed.

Step 4: Find the current in the $3.0\ \Omega$ parallel branch

This same p.d. of $6.0\text{ V}$ appears across both parallel branches: $I_{3.0\Omega} = \frac{V_p}{3.0} = \frac{6.0}{3.0} = 2.0\text{ A}$

Check using Kirchhoff’s first law: the current in the $6.0\ \Omega$ branch is $I_{6.0\Omega}=\frac{6.0}{6.0}=1.0\text{ A}$, and $2.0+1.0=3.0\text{ A}=I$, matching the total current found in Step 2. Confirmed.

Step 5: Why the other options are wrong

• B ($1.0\text{ A}$): this is the current in the other branch (the $6.0\ \Omega$ resistor), not the $3.0\ \Omega$ branch asked for.
• C ($3.0\text{ A}$): this is the total current supplied by the battery, as if it all flowed through the single $3.0\ \Omega$ parallel branch and none through the $6.0\ \Omega$ branch.
• D ($1.5\text{ A}$): this comes from wrongly splitting the total current equally between the two branches ($\frac{3.0}{2}$), ignoring that the branches have different resistances.

Final answer

• $V_p=6.0\text{ V}$ across the parallel section, so $I_{3.0\Omega}=\dfrac{6.0}{3.0}=\boxed{2.0}\text{ A}$, option A.