D.C. Circuits: Question 9

Worked solution

Part (a): Kirchhoff’s first law at junction $P$

Kirchhoff’s first law is a consequence of the conservation of charge: the total current flowing into a junction must equal the total current flowing out of it. At $P$, currents $I_1$ and $I_2$ flow in and current $I_3$ flows out, so: $I_1 + I_2 = I_3$

Part (b): Kirchhoff’s second law for each loop

Kirchhoff’s second law states that, around any complete loop, the sum of the e.m.f.s equals the sum of the potential drops. In both loops, current $I_3$ passes through the shared $3.0\ \Omega$ resistor.

Loop containing the $18\text{ V}$ battery: $18 = I_1(2.0) + I_3(3.0)$

Loop containing the $16\text{ V}$ battery: $16 = I_2(4.0) + I_3(3.0)$

Part (c): Solving the three equations simultaneously

Rearranging the two loop equations for $I_1$ and $I_2$: $I_1 = \frac{18-3.0I_3}{2.0} = 9.0 - 1.5I_3 \quad \text{...(1)}$ $I_2 = \frac{16-3.0I_3}{4.0} = 4.0 - 0.75I_3 \quad \text{...(2)}$

Substituting (1) and (2) into the junction equation $I_1+I_2=I_3$: $(9.0-1.5I_3) + (4.0-0.75I_3) = I_3$ $13.0 - 2.25I_3 = I_3$ $13.0 = 3.25I_3$ $I_3 = \frac{13.0}{3.25} = 4.0\text{ A}$

Substituting back into (1) and (2): $I_1 = 9.0 - 1.5(4.0) = 9.0-6.0 = 3.0\text{ A}$ $I_2 = 4.0 - 0.75(4.0) = 4.0-3.0 = 1.0\text{ A}$

Check using the junction equation: $I_1+I_2=3.0+1.0=4.0\text{ A}=I_3$. Confirmed.

Check by substituting back into the original loop equations:

• $18\text{ V}$ loop: $2.0(3.0)+3.0(4.0)=6.0+12.0=18.0\text{ V}$ ✓ matches the e.m.f.
• $16\text{ V}$ loop: $4.0(1.0)+3.0(4.0)=4.0+12.0=16.0\text{ V}$ ✓ matches the e.m.f.

Both loop equations and the junction equation are satisfied, confirming $I_1=3.0\text{ A}$, $I_2=1.0\text{ A}$, $I_3=4.0\text{ A}$.

Part (d): Potential difference across the $3.0\ \Omega$ resistor

$V_{3.0\Omega} = I_3 \times 3.0 = 4.0\times3.0 = 12\text{ V}$

Check using either loop: from the $18\text{ V}$ loop, the p.d. across the $2.0\ \Omega$ resistor is $I_1\times2.0=3.0\times2.0=6.0\text{ V}$, and $6.0+12.0=18.0\text{ V}$, matching the $18\text{ V}$ e.m.f. Confirmed.

Final answers

• (a) $I_1+I_2=\boxed{I_3}$
• (b) $18=2.0I_1+3.0I_3$; $16=4.0I_2+3.0I_3$
• (c) $I_1=\boxed{3.0}\text{ A}$, $I_2=\boxed{1.0}\text{ A}$, $I_3=\boxed{4.0}\text{ A}$
• (d) P.d. across $3.0\ \Omega$ resistor $=\boxed{12}\text{ V}$