Problem Statement
Analyze the circuit: Analyze the circuit: A parallel RLC circuit (ideal $L$, ideal $C$, real resistor $R$ in parallel) is driven by a current source $I = I_0\cos(\omega t)$. Find the impedance and anti-resonance frequency. Total admittance: $$Y = \frac{1}{R} + \frac{1}{i\omega L} + i\omega C = \frac{1}{R} + i\left(\omeg
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$Y = \frac{1}{R} + \frac{1}{i\omega L} + i\omega C = \frac{1}{R} + i\left(\omeg
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:
$$
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Step 4 — Find individual branch currents/voltages using current divider or voltage divider rules.
Step 5 — Power check: $\sum P_{\text{supplied}} = \sum P_{\text{dissipated}}$
Worked Calculation
Substituting all values with units:
For $\mathcal{E} = 12\,\text{V}$, $r = 1\,\Omega$, $R_1 = 3\,\Omega$, $R_2 = 6\,\Omega$ in parallel:
$$
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Answer
$$\boxed{I = \dfrac{\mathcal{E}}{R_{eq}+r}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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