D.C. Circuits: Question 6

Worked solution

Step 1: Recall the formula for resistors in parallel

For resistors connected in parallel, the reciprocal of the combined resistance equals the sum of the reciprocals of the individual resistances, a result that follows from Kirchhoff’s first law applied to the shared p.d. across each branch: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Step 2: Substitute the three equal resistances

All three resistors have resistance $4.0\ \Omega$: $\frac{1}{R} = \frac{1}{4.0} + \frac{1}{4.0} + \frac{1}{4.0} = \frac{3}{4.0} = 0.75\ \Omega^{-1}$

Check by adding as decimals: $0.250+0.250+0.250=0.750$, the same total. Confirmed.

Step 3: Invert to find $R$

$R = \frac{1}{0.75} = 1.33\ \Omega \ (\text{3 s.f.})$

Check: $R\times0.75=1.33\times0.75=1.00$, which matches $\frac{3}{4.0}\times R$ needing to equal $1$; substituting $R=4.0/3=1.333...$ directly confirms $R=\frac{4.0}{3}=1.33\ \Omega$.

Step 4: Why the other options are wrong

• B ($12.0\ \Omega$): this comes from adding the three resistances directly ($4.0+4.0+4.0$), which is the rule for resistors in series, not parallel.
• C ($0.75\ \Omega$): this is the sum of the reciprocals, $\frac{1}{R}$, left un-inverted rather than converted into the actual combined resistance $R$.
• D ($2.0\ \Omega$): this comes from treating the network as if only two of the three resistors combine (dividing $4.0\ \Omega$ by $2$) and ignoring the third resistor’s extra path for current.

Final answer

• $\dfrac{1}{R}=\dfrac{3}{4.0}=0.75\ \Omega^{-1}$, so $R=\dfrac{4.0}{3}=\boxed{1.33}\ \Omega$, option A.