DC Circuits
Ohm's law, Kirchhoff's voltage and current laws; the significance of internal resistance; series, parallel and series-parallel networks.
DC circuits, summary notes
- Ohm's law: V = I × R. Rearranged as I = V / R and R = V / I.
- Kirchhoff's current law: the sum of currents entering a node equals the sum leaving it. Charge is conserved.
- Kirchhoff's voltage law: around any closed loop, the sum of the EMFs equals the sum of the I·R drops. Energy is conserved.
- In series: current is the same everywhere, voltages add, and resistances add.
- In parallel: voltage is the same across each branch, currents add, and 1/R_T = Σ(1/R).
- ⚠ Exam trap: two equal resistors in parallel give half the value of one. Total parallel resistance is always less than the smallest branch.
- Ohm's law
- V = I × R
- Series resistance
- R_T = R₁ + R₂ + R₃ …
- Parallel resistance
- 1 / R_T = 1/R₁ + 1/R₂ + 1/R₃ …
- Two resistors in parallel
- R_T = (R₁ × R₂) / (R₁ + R₂)
Two 10 Ω resistors in parallel are in series with a 5 Ω resistor. Find the total resistance.
Parallel pair = (10 × 10)/(10 + 10) = 5 Ω. In series with 5 Ω gives R_T = 10 Ω.
A 24 V supply drives 3 A through a circuit. What is its resistance?
R = V / I = 24 / 3 = 8 Ω.
Ohm's law, V = I × R
Series vs parallel resistance
Kirchhoff's laws
The voltage divider
DC circuits concept map
DC Circuits
DC circuits quiz
DC Circuits, quiz
1. For the circuit shown, the current drawn from the battery is:
2 A6 A0.5 A2. In the circuit shown, the voltage dropped across R2 is:
8 V4 V12 V3. For the parallel circuit shown, the total resistance is:
2 Ω9 Ω4.5 Ω4. In the parallel circuit shown, the current through the 3 Ω branch is:
8 A4 A12 A5. For the series-parallel circuit shown, the total resistance is:
8 Ω22 Ω6 Ω6. In the potential divider shown, the output voltage across R2 is:
5 V15 V6.7 V7. At the junction shown, 4 A and 3 A flow in and 5 A flows out on one branch. The current I is:
2 A12 A6 A8. The Wheatstone bridge shown is balanced, so the galvanometer reads zero. This means:
P/Q = R/SP + Q = R + SP × S = Q × R only when the supply is DC9. For the circuit shown, the current drawn from the supply is:
2 A20 A0.5 A10. The three resistors shown are identical and in parallel. The total resistance is:
4 Ω36 Ω12 Ω11. In the circuit shown, meter 1 is in series with the load and meter 2 is connected across it. Meters 1 and 2 are respectively:
An ammeter and a voltmeterA voltmeter and an ammeterBoth ammeters12. The symbol shown, a resistor with an arrow drawn through it, represents:
A variable resistorA fixed resistorA fuse
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