Problem Statement
Analyze the circuit: Analyze the circuit: Explain why doubling light intensity doubles photocurrent but does not change stopping voltage. Stopping voltage is set by $eV_s = h\nu – \phi$ — it depends on photon energy (frequency), not on the number of photons. Intensity determines the photon flux: double the intensity, do
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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