D.C. Circuits: Question 5

Worked solution

Step 1: Recall Kirchhoff’s second law

Kirchhoff’s second law is a consequence of the conservation of energy: going around any complete loop, the sum of the e.m.f.s equals the sum of the $IR$ potential drops. When two e.m.f.s in a loop oppose each other, they must be combined with opposite signs, giving a single net e.m.f. driving the current.

Step 2: Find the net e.m.f. of the loop

The two cells oppose each other, so the net e.m.f. is the difference between them, not their sum: $E_{\text{net}} = 6.0 - 2.0 = 4.0\text{ V}$

Check by reasoning about direction: the $6.0\text{ V}$ cell drives current one way around the loop, and the $2.0\text{ V}$ cell tries to drive current the opposite way; the stronger cell wins, but only by the amount left over after the weaker cell’s opposition, which is exactly $6.0-2.0=4.0\text{ V}$.

Step 3: Apply Kirchhoff’s second law to find the current

With both cells having negligible internal resistance, the only resistance in the loop is the $4.0\ \Omega$ resistor: $E_{\text{net}} = IR \implies I = \frac{E_{\text{net}}}{R} = \frac{4.0}{4.0} = 1.0\text{ A}$

Check: $1.0\times4.0=4.0\text{ V}$, which matches $E_{\text{net}}$. Confirmed.

Step 4: Why the other options are wrong

• B ($2.0\text{ A}$): comes from wrongly adding the e.m.f.s ($6.0+2.0=8.0\text{ V}$) instead of subtracting them, then dividing by $4.0\ \Omega$.
• C ($1.5\text{ A}$): comes from using only the larger cell’s e.m.f. ($6.0\text{ V}$) and ignoring the opposing $2.0\text{ V}$ cell, then dividing by $4.0\ \Omega$.
• D ($0.5\text{ A}$): comes from using only the smaller cell’s e.m.f. ($2.0\text{ V}$) and ignoring the larger cell, then dividing by $4.0\ \Omega$.

Final answer

• Net e.m.f. $=6.0-2.0=4.0\text{ V}$, so $I=\dfrac{4.0}{4.0}=\boxed{1.0}\text{ A}$, option A.