Series parallel circuits and Kirchhoff rules

A mixed circuit contains resistors, some of which are connected in series and others in parallel. A series section gives $R_{\text{s}}=\sum_i R_i$, a parallel section gives $1/R_{\text{p}}=\sum_i 1/R_i$, so the whole network can be reduced step by step to a single equivalent resistance. When several emf sources sit in different branches, Ohm's law alone is not enough, and Kirchhoff's rules provide the complete system of equations. The junction rule states that the sum of the currents entering a node equals the sum of the currents leaving it: $\sum I_{\text{in}}=\sum I_{\text{out}}$. The loop rule states that along a closed loop the algebraic sum of the emf sources equals the algebraic sum of the $IR$ drops: $\sum \varepsilon=\sum IR$. Before applying the rules, current directions and loop traversal directions are chosen arbitrarily, and a negative answer merely means that the real current flows in the opposite direction.

A circuit consists of a source $\varepsilon=20\ \text{V}$ (internal resistance neglected), a resistor $R_1=4\ \Omega$, a parallel section made of $R_2=12\ \Omega$ and $R_3=6\ \Omega$, and a resistor $R_4=2\ \Omega$, all connected in series. Find the equivalent resistance, the total current, the voltage across the parallel section, and the currents in the branches $R_2$ and $R_3$.

Node A is connected to node B by three branches. Left branch: source $\varepsilon_1=12\ \text{V}$ (positive terminal toward node A) and $R_1=2\ \Omega$. Middle branch: $R_2=4\ \Omega$. Right branch: $R_3=2\ \Omega$ and source $\varepsilon_2=6\ \text{V}$ (positive terminal toward node A). Let $I_1$ be the current in the left branch upward (B → A), $I_2$ the current in the middle branch downward (A → B), and $I_3$ the current in the right branch downward (A → B). Find all three currents.

Take the same circuit as in the previous example ($\varepsilon_1=12\ \text{V}$, $R_1=2\ \Omega$, $R_2=4\ \Omega$, $R_3=2\ \Omega$, source positive terminal toward node A), but replace the right source by $\varepsilon_2=15\ \text{V}$ with the same polarity. Determine the direction of the current in the right branch, decide whether this source delivers or absorbs energy, and verify the result with a power balance.

A safe self-study plan in a simulator (PhET "Circuit Construction Kit: DC" or an equivalent; no live wiring at voltage is needed here).

1. Build the circuit of Example 2: two sources, three branches between nodes A and B, an ammeter in each branch.
2. Record the ammeter readings and check the junction rule: the sum of currents entering A must equal the sum leaving it.
3. Connect a voltmeter across each branch in turn and confirm that $V_A-V_B$ is the same for all three branches.
4. Add up the $IR$ drops and emfs around the left and right loops and confirm that both sums equal zero.
5. Slowly vary $\varepsilon_2$ from $+6\ \text{V}$ to $-18\ \text{V}$ (negative values mean the right source has the opposite polarity to the original one) and find the value at which the current in the middle branch becomes zero; this occurs at $\varepsilon_2=-12\ \text{V}$ in the original sign convention. Record this threshold and explain it using the Kirchhoff equations: $I_2=(12+\varepsilon_2)/10\ \text{A}$.
6. Compare the measured currents with the values computed in Examples 2 and 3; a discrepancy above 5 % means an error in signs or in the chosen loop direction.

To show the topology of the mixed circuit and explicitly mark the node and the loops to which Kirchhoff's rules are applied, linking the schematic to the system of equations.

A schematic sketch: two nodes A (top) and B (bottom); the left branch with $\varepsilon_1$ and $R_1$, the middle branch with $R_2$, the right branch with $R_3$ and $\varepsilon_2$; on each branch an arrow shows the chosen positive current direction ($I_1$ upward, $I_2$ and $I_3$ downward); dashed circulation arrows mark the left and right loops, and at node A a short arc marks the junction rule $I_1=I_2+I_3$.

A bottle dynamo on a bicycle delivers $6.0\ \text{V}$. The headlight with resistance $R_1 = 12\ \Omega$ and the tail light with $R_2 = 24\ \Omega$ are connected in parallel. The connecting wire has a total resistance $R_w = 0.4\ \Omega$ and is in series with that pair. While riding, the headlight glows brightly and the small tail light only dimly.

The two lamps form a parallel section: $1/R_p = 1/12 + 1/24$, hence $R_p = 8.0\ \Omega$. The wire adds in series with it: $R = 8.0 + 0.4 = 8.4\ \Omega$, so $I = 6.0/8.4 \approx 0.71\ \text{A}$. The drop along the wire is $0.4 \cdot 0.71 \approx 0.29\ \text{V}$, so the node sits at $5.71\ \text{V}$. Kirchhoff's current rule at the junction: $0.48\ \text{A} + 0.24\ \text{A} \approx 0.71\ \text{A}$. The loop rule: $6.0 = 0.29 + 5.71$.

A power bank outputs $5.00\ \text{V}$. The USB cable has a total resistance $R_c = 0.15\ \Omega$ and carries $I = 2.0\ \text{A}$. Inside the phone the current splits: $1.7\ \text{A}$ flows into the battery while $0.3\ \text{A}$ runs the screen and the processor.

The cable and the phone are in series, while inside the phone the battery and the running electronics form parallel branches. By the loop rule the cable drop is $0.15 \cdot 2.0 = 0.30\ \text{V}$, so the phone input sees $5.00 - 0.30 = 4.70\ \text{V}$; the cable turns $I^2 R_c = 4.0 \cdot 0.15 = 0.60\ \text{W}$ into heat. By the current rule at the node: $2.0 = 1.7 + 0.3$.

In a kitchen the socket circuit is fed at $230\ \text{V}$ through cable of resistance $R_w = 0.10\ \Omega$ and protected by a $16\ \text{A}$ breaker. The kettle is rated $2.0\ \text{kW}$ (current $\approx 8.7\ \text{A}$) and the microwave $1.0\ \text{kW}$ (current $\approx 4.3\ \text{A}$). They are switched on together.

The appliances are parallel branches, and the cable is in series with both. Current rule: at the junction the line current is $8.7 + 4.3 = 13.0\ \text{A}$, below the $16\ \text{A}$, so the breaker holds. Voltage rule: the cable drop is $0.10 \cdot 13.0 = 1.3\ \text{V}$, and each appliance receives $228.7\ \text{V}$ instead of $230\ \text{V}$. With only the kettle running, the $8.7\ \text{A}$ current causes a drop of just $0.87\ \text{V}$: adding a parallel branch increases the drop across the series part upstream of it.