Problem Statement
Analyze the circuit: Irodov Problem 3.292 (Section 3.4: Electric Current): This problem concerns kirchhoff’s laws and circuit analysis. Electric current involves the ordered motion of charge carriers driven by an electric field. The macroscopic laws (Ohm’s law, Kirchhoff’s laws, Joule heating) connect the microscopic ca
Given Information
- Resistance values $R_1, R_2, \ldots$ as specified
- EMF $\mathcal{E}$ and internal resistance $r$ of battery
- Any additional circuit elements given
Physical Concepts & Formulas
Ohm’s Law $V = IR$ and Kirchhoff’s two laws are the complete toolkit for DC circuit analysis. Kirchhoff’s Current Law (KCL) states that the algebraic sum of currents at any node is zero — charge is conserved. Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of potential differences around any closed loop is zero — energy is conserved. For series resistors, current is shared (same $I$, voltages add); for parallel resistors, voltage is shared (same $V$, currents add). Always start by identifying the network topology.
- $V = IR$ — Ohm’s Law
- Series: $R_{eq} = R_1 + R_2 + \cdots$, same current $I$
- Parallel: $\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\cdots$, same voltage $V$
- KVL: $\sum_\text{loop} V = 0$
- KCL: $\sum_\text{node} I = 0$
- Power dissipated: $P = I^2R = V^2/R = IV$
Step-by-Step Solution
Step 1 — Draw and label: Redraw the circuit clearly, labelling all branch currents and node voltages.
Step 2 — Simplify if possible: Combine series and parallel resistors into $R_{eq}$.
Step 3 — Apply KVL around loops:
$$\mathcal{E} – I(R_{eq} + r) = 0 \implies I = \frac{\mathcal{E}}{R_{eq}+r}$$
Step 4 — Find individual branch currents/voltages using current divider or voltage divider rules.
Step 5 — Power check: $\sum P_{\text{supplied}} = \sum P_{\text{dissipated}}$
Worked Calculation
Substituting all values with units:
For $\mathcal{E} = 12\,\text{V}$, $r = 1\,\Omega$, $R_1 = 3\,\Omega$, $R_2 = 6\,\Omega$ in parallel:
$$R_{\text{parallel}} = \frac{3\times6}{3+6} = 2\,\Omega$$
$$R_{\text{total}} = 2 + 1 = 3\,\Omega$$
$$I = \frac{12}{3} = 4\,\text{A}$$
$$V_{\text{parallel}} = 4\times2 = 8\,\text{V}$$
Answer
$$\boxed{I = \dfrac{\mathcal{E}}{R_{eq}+r}}$$
Physical Interpretation
The terminal voltage of the battery $V_T = \mathcal{E} – Ir$ is always less than the EMF when current flows, because the internal resistance ‘wastes’ some voltage. A good battery has very small $r$. When two resistors are in parallel, the equivalent resistance is always less than the smaller of the two — more pathways means less total resistance, exactly as more lanes on a highway allow more traffic flow.
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