Problem Statement
Analyze the circuit: Analyze the circuit: A series RLC circuit ($R = 10$ $\Omega$, $L = 35$ mH, $C = 80$ $\mu$F) is driven by $\mathcal{E}_0 = 100$ V. Find the current at resonance and the $Q$-factor. $$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{35\times10^{-3}\times80\times10^{-6}}} = \frac{1}{\sqrt{2.8\times10^{-
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.
- $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
- Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
- $f_k = \mu_k N$ — kinetic friction
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{35\times10^{-3}\times80\times10^{-6}}} = \frac{1}{\sqrt{2.8\times10^{-
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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