DC Circuits -- Diagnostic Tests

flowchart TD
    A[Diag Dc Circuits] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Physics describes the fundamental rules of the universe, from the tiniest particles to the vastness of space.

Question:

A battery has EMF $\varepsilon = 12.0\,\text{V}$ and internal resistance $r = 2.0\,\Omega$. It is connected to a variable external resistance $R$.

(a) Derive an expression for the power delivered to $R$ and find the value of $R$ that maximises this power.

(b) Calculate the maximum power delivered to $R$ and the efficiency at this operating point.

(c) Calculate the power delivered to $R$ and the efficiency when $R = 10\,\Omega$.

Solution:

(a) Current: $I = \varepsilon/(R + r)$

Power to $R$: $P = I^2R = \frac{\varepsilon^2 R}{(R + r)^2}$

To maximise: $\frac{dP}{dR} = \frac{\varepsilon^2(R + r)^2 - \varepsilon^2 R \times 2(R + r)}{(R + r)^4} = 0$

$\varepsilon^2(R + r)[(R + r) - 2R] = 0$ $R + r - 2R = 0 \Rightarrow R = r = 2.0\,\Omega$

Maximum power transfer occurs when the load resistance equals the internal resistance.

(b) At $R = r = 2.0\,\Omega$:

$I = 12.0/4.0 = 3.0\,\text{A}$ $P_{\max} = I^2R = 9.0 \times 2.0 = 18.0\,\text{W}$

Efficiency $= P_R/P_{\text{total}} = I^2R/(I^2(R + r)) = R/(R + r) = 2.0/4.0 = 50\%$

Half the power is dissipated in the internal resistance.

(c) At $R = 10\,\Omega$:

$I = 12.0/12.0 = 1.0\,\text{A}$ $P = 1.0 \times 10 = 10.0\,\text{W}$

Efficiency $= 10.0/12.0 = 83.3\%$

Higher efficiency but lower power than the matched condition. This is the fundamental trade-off in power transfer.

Question:

A potential divider consists of two resistors $R_1 = 10\,\text{k}\Omega$ and $R_2 = 20\,\text{k}\Omega$ in series across a $15.0\,\text{V}$ supply with negligible internal resistance. The output is taken across $R_2$.

(a) Calculate the output voltage with no load connected.

(b) A load resistor $R_L = 10\,\text{k}\Omega$ is connected across $R_2$. Calculate the new output voltage.

(c) What value of $R_L$ would reduce the output voltage to $50\%$ of the unloaded value? Comment on your answer.

Solution:

(a) Unloaded: $V_{\text{out}} = V_{\text{in}} \times \frac{R_2}{R_1 + R_2} = 15.0 \times \frac{20}{30} = 10.0\,\text{V}$

(b) With load, $R_2$ and $R_L$ are in parallel:

The output voltage drops from $10.0\,\text{V}$ to $6.0\,\text{V}$ — a $40\%$ reduction. This “loading effect” occurs because the load draws current, changing the effective resistance of the divider.

(c) We need $V_{\text{out}} = 5.0\,\text{V}$ (50% of $10.0\,\text{V}$):

$200 + 30 R_L = 60 R_L$ $200 = 30 R_L$ $R_L = 6.67\,\text{k}\Omega$

When the load resistance equals the parallel combination resistance, the output drops to half. This demonstrates why potential dividers should have $R_L \gg R_2$ to avoid loading effects.

Question:

Two cells with EMFs $\varepsilon_1 = 10.0\,\text{V}$ (internal resistance $r_1 = 1.0\,\Omega$) and $\varepsilon_2 = 6.0\,\text{V}$ (internal resistance $r_2 = 2.0\,\Omega$) are connected in parallel across a resistor $R = 5.0\,\Omega$. The positive terminals are connected together.

(a) Calculate the current through each cell and through $R$.

(b) Calculate the terminal potential difference across each cell.

(c) Calculate the power delivered by each cell and the power dissipated in $R$. Verify conservation of energy.

Solution:

(a) Let $I_1$ flow from $\varepsilon_1$ (out of positive terminal), $I_2$ flow from $\varepsilon_2$ And $I_R$ flow through $R$ (downward). Apply Kirchhoff’s current law at the top junction:

Apply Kirchhoff’s voltage law to the left loop ($\varepsilon_1$, $r_1$, $R$):

Apply Kirchhoff’s voltage law to the right loop ($\varepsilon_2$, $r_2$, $R$):

From (2): $I_2 = (6.0 - 5I_R)/2 = 3.0 - 2.5I_R$

From (1): $I_1 = 10.0 - 5I_R$

Substituting into KCL: $(10.0 - 5I_R) + (3.0 - 2.5I_R) = I_R$

$13.0 - 7.5I_R = I_R$ $13.0 = 8.5I_R$ $I_R = 1.529\,\text{A}$

The negative sign for $I_2$ means $\varepsilon_2$ is being charged by $\varepsilon_1$. Current flows into the positive terminal of $\varepsilon_2$.

(b) Terminal p.d. Across $\varepsilon_1$: $V_1 = \varepsilon_1 - I_1 r_1 = 10.0 - 2.353 = 7.647\,\text{V}$

Terminal p.d. Across $\varepsilon_2$: $V_2 = \varepsilon_2 - I_2 r_2 = 6.0 - (-0.824)(2.0) = 6.0 + 1.647 = 7.647\,\text{V}$

Both terminal p.d.s are equal, as expected since they are connected in parallel.

(c) Power from $\varepsilon_1$: $P_1 = \varepsilon_1 I_1 = 10.0 \times 2.353 = 23.53\,\text{W}$

Power absorbed by $\varepsilon_2$: $P_2 = \varepsilon_2 I_2 = 6.0 \times 0.824 = 4.94\,\text{W}$ (charging)

Power dissipated in $R$: $P_R = I_R^2 R = (1.529)^2 \times 5.0 = 11.69\,\text{W}$

Power dissipated in $r_1$: $I_1^2 r_1 = (2.353)^2 \times 1.0 = 5.54\,\text{W}$

Power dissipated in $r_2$: $I_2^2 r_2 = (0.824)^2 \times 2.0 = 1.36\,\text{W}$

Conservation check: $P_1 = P_R + P_{r1} + P_{r2} + P_2$ (absorbed by $\varepsilon_2$)

$23.53 = 11.69 + 5.54 + 1.36 + 4.94 = 23.53$. Conserved.

Question:

A solar panel produces an EMF of $18.0\,\text{V}$ and has internal resistance $2.0\,\Omega$. It is used to charge a rechargeable battery of EMF $12.0\,\text{V}$ and internal resistance $1.0\,\Omega$. A capacitor of capacitance $1000\,\mu\text{F}$ is connected in parallel with the battery to smooth the charging current.

(a) Calculate the initial charging current.

(b) Calculate the power delivered by the solar panel and the power absorbed by the battery at the start of charging.

(c) The capacitor is initially uncharged. Calculate the time constant for the charging of the capacitor and the energy stored in the capacitor when fully charged.

Solution:

(a) The capacitor is in parallel with the battery, so the capacitor charges to the battery terminal voltage. The steady-state charging current is determined by the series circuit of solar panel, internal resistances, and battery:

In steady state (capacitor fully charged):

$I = \frac{18.0 - (12.0 + I)}{2.0}$ $2I = 6.0 - I$ $3I = 6.0 \Rightarrow I = 2.0\,\text{A}$

The steady-state charging current is $2.0\,\text{A}$. Initially, the uncharged capacitor draws additional transient current, but this decays exponentially.

(b) Power from panel: $P_{\text{panel}} = \varepsilon_{\text{panel}} \times I = 18.0 \times 2.0 = 36.0\,\text{W}$

Power to battery: $P_{\text{battery}} = \varepsilon_{\text{battery}} \times I = 12.0 \times 2.0 = 24.0\,\text{W}$

Power dissipated in internal resistances: $P_{\text{loss}} = I^2(r_1 + r_2) = 4.0 \times 3.0 = 12.0\,\text{W}$

Check: $36.0 = 24.0 + 12.0$. Consistent.

(c) The capacitor charges through the internal resistance of the battery: $\tau = R_{\text{battery}} \times C = 1.0 \times 1000 \times 10^{-6} = 1.0 \times 10^{-3}\,\text{s} = 1.0\,\text{ms}$

The capacitor charges to the battery terminal voltage: $V_C = 12.0 + 2.0 \times 1.0 = 14.0\,\text{V}$

Energy stored: $E = \frac{1}{2}CV^2 = 0.5 \times 10^{-3} \times 196 = 0.098\,\text{J}$

Question:

A circuit consists of a $24\,\text{V}$ supply (internal resistance $1\,\Omega$) in series with $R_1 = 5\,\Omega$ Which is then connected to a parallel combination of $R_2 = 10\,\Omega$ and $R_3 = 15\,\Omega$.

(a) Find the Thevenin equivalent circuit across $R_3$.

(b) Using the Thevenin equivalent, calculate the current through $R_3$.

(c) $R_3$ is replaced by a variable resistor. Calculate the range of values of $R_3$ for which the current through it exceeds $0.8\,\text{A}$.

Solution:

(a) Remove $R_3$ and find the open-circuit voltage across the terminals:

The current through $R_1$ and $R_2$ in series (with $R_3$ removed):

Open-circuit voltage: $V_{\text{OC}} = 24 - 1.5(1 + 5) = 24 - 9 = 15\,\text{V}$

Or: $V_{\text{OC}} = I \times R_2 = 1.5 \times 10 = 15\,\text{V}$. Consistent.

Internal resistance (replace the $24\,\text{V}$ source with a short circuit, and remove $R_3$):

$r = 1\,\Omega$ and $R_1 = 5\,\Omega$ are in series $= 6\,\Omega$. This combination is in parallel with $R_2 = 10\,\Omega$.

Thevenin equivalent: $V_{\text{Th}} = 15\,\text{V}$, $R_{\text{Th}} = 3.75\,\Omega$.

(b) Current through $R_3$:

(c) We need $I_3 \gt 0.8\,\text{A}$:

$\frac{15}{3.75 + R_3} \gt 0.8$ $15 \gt 0.8(3.75 + R_3)$ $18.75 \gt 3.0 + 0.8 R_3$ $0.8 R_3 \lt 15.75$ $R_3 \lt 19.7\,\Omega$

For any $R_3 \lt 19.7\,\Omega$ The current exceeds $0.8\,\text{A}$. Since the original $R_3 = 15\,\Omega$ gives exactly $0.8\,\text{A}$ This is the boundary.

Question:

A circuit contains a resistor $R = 47\,\Omega$ connected to a $9.0\,\text{V}$ supply. A student uses a non-ideal ammeter (resistance $R_A = 0.50\,\Omega$) and a non-ideal voltmeter (resistance $R_V = 5000\,\Omega$) to measure the current through and voltage across $R$.

(a) Calculate the current and voltage across $R$ if both meters are ideal.

(b) Calculate the measured current when the ammeter is placed in series with $R$.

(c) Calculate the measured voltage when the voltmeter is placed in parallel with $R$. By what percentage does each measurement differ from the ideal value?

Solution:

(a) Ideal: $I = V/R = 9.0/47 = 0.1915\,\text{A}$; $V_R = 9.0\,\text{V}$ (across $R$ alone).

(b) With ammeter in series: total resistance $= R + R_A = 47.5\,\Omega$

Percentage error: $\frac{0.1915 - 0.1895}{0.1915} \times 100 = 1.04\%$

The ammeter under-reads by about $1\%$ because it adds its own resistance to the circuit.

(c) With voltmeter in parallel: $R \parallel R_V = \frac{47 \times 5000}{47 + 5000} = \frac{235000}{5047} = 46.56\,\Omega$

The voltmeter draws current, reducing the voltage across $R$. With an ideal supply (negligible internal resistance) and only the parallel combination as the load:

The voltmeter reads $9.0\,\text{V}$ Which equals the ideal value. This is because with no other resistance in the circuit, the full supply voltage appears across the parallel combination regardless of the voltmeter’s loading effect.

However, if the supply has internal resistance $r = 5\,\Omega$:

Without voltmeter: $V_R = 9.0 \times 47/(5 + 47) = 9.0 \times 47/52 = 8.134\,\text{V}$

With voltmeter: total load $= 46.56\,\Omega$

Percentage error: $(8.134 - 8.127)/8.134 \times 100 = 0.086\%$

The voltmeter has negligible effect because $R_V \gg R$.

Confusing current and voltage: Current (A) is the flow of charge; voltage (V) is the energy per unit charge. Resistance opposes current, not voltage. Students often say “the resistor uses up the voltage” when they should say “the voltage drops across the resistor.”

Forgetting Ohm’s law applies only to ohmic conductors: = IR$is valid only when resistance is constant (ohmic conductors at constant temperature). For non-ohmic devices (diodes, thermistors, filament lamps), resistance changes with voltage or temperature, so = IR$ gives the resistance at that point, not a constant.

Confusing the conventions for electron flow and conventional current: Conventional current flows from positive to negative (the direction a positive charge would move). Electron flow is from negative to positive (the actual movement of electrons). Most circuit analysis uses conventional current. Using electron flow when conventional current is expected gives reversed directions.

• Mechanics: Mechanics covers motion, forces, and energy
• Waves: Waves transfer energy through oscillations
• Electricity: Electricity covers circuits and fields