Solved Examples

Solved Examples

A dedicated bank of worked problems. Core formulae:

• Current: $I = \dfrac{Q}{t}$ · Potential difference: $V = \dfrac{W}{Q}$ · Ohm's law: $V = IR$
• Resistance: $R = \rho \dfrac{l}{A}$ · Series: $R_s = R_1 + R_2 + \dots$ · Parallel: $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$
• Power: $P = VI = I^2R = \dfrac{V^2}{R}$ · Heat: $H = I^2Rt$ · Energy: $1\ \text{kW h} = 3.6 \times 10^6\ \text{J}$

Example 1: Charge (NCERT 11.1) Current 0.5 A for 10 min. Find the charge. Solution: $Q = It = 0.5 \times 600 = 300\ \text{C}$.

Example 2: Work done (NCERT 11.2) Move 2 C across 12 V. Find the work. Solution: $W = VQ = 12 \times 2 = 24\ \text{J}$.

Example 3: Current drawn by bulb and heater (NCERT 11.3) 220 V source. (a) bulb R = 1200 Ω, (b) heater R = 100 Ω. Solution: (a) $I = 220/1200 = 0.18\ \text{A}$. (b) $I = 220/100 = 2.2\ \text{A}$.

Example 4: Heater at higher voltage (NCERT 11.4) 60 V gives 4 A. Find the current at 120 V. Solution: $R = 60/4 = 15\ \Omega$; at 120 V, $I = 120/15 = 8\ \text{A}$.

Example 5: Resistivity (NCERT 11.5) R = 26 Ω, l = 1 m, d = 0.3 mm. Find resistivity. Solution: $\rho = \dfrac{R\pi d^2}{4l} = 1.84 \times 10^{-6}\ \Omega\,\text{m}$ (manganese).

Example 6: Reshaping a wire (NCERT 11.6) R = 4 Ω. New wire: length l/2, area 2A. Find R. Solution: $R_2 = \rho\dfrac{l/2}{2A} = \dfrac{1}{4}(4) = 1\ \Omega$.

Example 7: Lamp and conductor in series (NCERT 11.7) 20 Ω lamp + 4 Ω conductor on 6 V. Solution: $R_s = 24\ \Omega$; $I = 6/24 = 0.25\ \text{A}$; $V_1 = 5\ \text{V}$, $V_2 = 1\ \text{V}$.

Example 8: Three resistors in parallel (NCERT 11.8) 5 Ω, 10 Ω, 30 Ω on 12 V. Solution: $I_1 = 2.4$, $I_2 = 1.2$, $I_3 = 0.4$ A; $I = 4\ \text{A}$; $R_p = 3\ \Omega$.

Example 9: Series-parallel network (NCERT 11.9) R1=10, R2=40 (parallel); R3=30, R4=20, R5=60 (parallel); groups in series; 12 V. Solution: $R' = 8\ \Omega$, $R'' = 10\ \Omega$, total $R = 18\ \Omega$; $I = 12/18 = 0.67\ \text{A}$.

Example 10: Iron - current and resistance (NCERT 11.10) 840 W and 360 W at 220 V. Solution: Max: $I = 3.82\ \text{A}$, $R = 57.6\ \Omega$. Min: $I = 1.64\ \text{A}$, $R = 134.15\ \Omega$.

Example 11: PD from heating (NCERT 11.11) 100 J/s in 4 Ω. Find V. Solution: $I = \sqrt{H/Rt} = \sqrt{100/4} = 5\ \text{A}$; $V = IR = 20\ \text{V}$.

Example 12: Power of a bulb (NCERT 11.12) 220 V, 0.50 A. Solution: $P = VI = 220 \times 0.5 = 110\ \text{W}$.

Example 13: Cost of energy (NCERT 11.13) 400 W, 8 h/day, 30 days at Rs 3.00/kW h. Solution: Energy $= 96\ \text{kW h}$; cost $= 96 \times 3 = \text{Rs}\ 288$.

Example 14: Wire cut into 5 parts (NCERT Ex Q1) A wire of resistance R is cut into 5 equal parts, joined in parallel. Find R/R'. Solution: Each part $= R/5$. Five in parallel: $R' = \dfrac{R/5}{5} = \dfrac{R}{25}$. So $\dfrac{R}{R'} = 25$.

Example 15: Bulb at half voltage (NCERT Ex Q3) 220 V, 100 W bulb operated on 110 V. Solution: $R = V^2/P = 484\ \Omega$; at 110 V, $P = 110^2/484 = 25\ \text{W}$.

Example 16: Unknown resistor (NCERT Ex Q8) 12 V battery gives 2.5 mA. Find R. Solution: $R = V/I = 12/(2.5 \times 10^{-3}) = 4800\ \Omega = 4.8\ \text{k}\Omega$.

Example 17: Current through 12 Ω (NCERT Ex Q9) 9 V in series with 0.2, 0.3, 0.4, 0.5, 12 Ω. Solution: $R_s = 13.4\ \Omega$; $I = 9/13.4 = 0.67\ \text{A}$ (same through the 12 Ω).

Example 18: Number of parallel resistors (NCERT Ex Q10) How many 176 Ω resistors in parallel carry 5 A on 220 V? Solution: Needed $R = V/I = 220/5 = 44\ \Omega$. For $n$ equal resistors, $R = 176/n = 44$, so $n = 4$. Four resistors.

Example 19: Number of 10 W lamps (NCERT Ex Q12) 10 W lamps on 220 V, max allowable current 5 A. How many in parallel? Solution: Total power $= VI = 220 \times 5 = 1100\ \text{W}$. Number $= 1100/10 = 110$ lamps.

Example 20: Hot plate coils (NCERT Ex Q13) Two 24 Ω coils on 220 V: used (a) separately, (b) in series, (c) in parallel. Find currents. Solution: (a) $I = 220/24 = 9.17\ \text{A}$. (b) series $R = 48\ \Omega$, $I = 220/48 = 4.58\ \text{A}$. (c) parallel $R = 12\ \Omega$, $I = 220/12 = 18.33\ \text{A}$.

Example 21: Current drawn by two lamps (NCERT Ex Q15) 100 W and 60 W lamps (both 220 V) in parallel on 220 V mains. Solution: Total power $= 100 + 60 = 160\ \text{W}$. $I = P/V = 160/220 = 0.73\ \text{A}$.

Example 22: Which uses more energy? (NCERT Ex Q16) 250 W TV for 1 h vs 1200 W toaster for 10 min. Solution: TV $= 250 \times 1 = 250\ \text{W h}$. Toaster $= 1200 \times (10/60) = 200\ \text{W h}$. The TV uses more energy.

Example 23: Rate of heat in a heater (NCERT Ex Q17) Heater R = 44 Ω draws 5 A for 2 h. Find the rate of heat developed. Solution: Rate of heat = power $= I^2R = 5^2 \times 44 = 25 \times 44 = 1100\ \text{W} = 1100\ \text{J/s}$.

Example 24: Length of a copper wire (NCERT Ex Q6) Copper wire d = 0.5 mm, $\rho = 1.6 \times 10^{-8}$ Ω m, R = 10 Ω. Find l. What if the diameter is doubled? Solution: $A = \pi d^2/4 = \pi(5 \times 10^{-4})^2/4 = 1.96 \times 10^{-7}\ \text{m}^2$. $l = RA/\rho = (10 \times 1.96 \times 10^{-7})/(1.6 \times 10^{-8}) \approx 122.7\ \text{m}$. If the diameter is doubled, area becomes 4 times, so resistance becomes one-quarter (2.5 Ω).

Example 25: Power in a 2 Ω resistor (NCERT Ex Q14) (i) 6 V battery in series with 1 Ω and 2 Ω. (ii) 4 V battery in parallel with 12 Ω and 2 Ω. Compare power in the 2 Ω resistor. Solution: (i) $I = 6/3 = 2\ \text{A}$; $P = I^2R = 4 \times 2 = 8\ \text{W}$. (ii) 2 Ω has 4 V across it; $P = V^2/R = 16/2 = 8\ \text{W}$. Both are 8 W.

Example 26: Heat while transferring charge (NCERT Q) Transfer 96000 C in 1 hour through 50 V. Find the heat. Solution: $H = VQ = 50 \times 96000 = 4.8 \times 10^6\ \text{J}$ (time is not needed since H = VIt = VQ).

Example 27: Motor power and energy (NCERT Q) A motor takes 5 A from a 220 V line. Find its power and the energy used in 2 h. Solution: $P = VI = 220 \times 5 = 1100\ \text{W} = 1.1\ \text{kW}$. Energy $= 1.1 \times 2 = 2.2\ \text{kW h} = 7.92 \times 10^6\ \text{J}$.

Example 28: Highest and lowest resistance (NCERT Q) Four coils 4, 8, 12, 24 Ω. Find (a) highest, (b) lowest possible total resistance. Solution: (a) All in series: $4+8+12+24 = 48\ \Omega$. (b) All in parallel: $\dfrac{1}{R} = \dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24} = \dfrac{6+3+2+1}{24} = \dfrac{12}{24}$, so $R = 2\ \Omega$.

Example 29: Equivalent resistance quick checks (NCERT Q) Find the equivalent resistance of (a) 1 Ω and 10⁶ Ω in parallel, (b) 1 Ω, 10³ Ω and 10⁶ Ω in parallel. Solution: In parallel the result is just below the smallest resistance. (a) $\approx 1\ \Omega$ (the huge resistor carries almost no current). (b) $\approx 1\ \Omega$ for the same reason.

Example 30: Resistance of an equivalent iron (NCERT Q) A 100 Ω lamp, 50 Ω toaster and 500 Ω filter are in parallel on 220 V. Find the single equivalent iron resistance drawing the same total current, and that current. Solution: $\dfrac{1}{R_p} = \dfrac{1}{100} + \dfrac{1}{50} + \dfrac{1}{500} = \dfrac{5+10+1}{500} = \dfrac{16}{500}$, so $R_p = 31.25\ \Omega$. Current $= 220/31.25 = 7.04\ \text{A}$.

Example 31: Heat ratio series vs parallel (NCERT Ex Q4) Two identical wires connected first in series then in parallel across the same voltage. Find the ratio of heat produced. Solution: Series $R_s = 2R$, parallel $R_p = R/2$. For the same $V$ and time, $H = V^2t/R$, so $H_{series}:H_{parallel} = \dfrac{1}{2R} : \dfrac{2}{R} = 1 : 4$.

Example 32: Reading R from a V-I table (NCERT Ex Q7) For a resistor, when V = 3.4 V the current is 1.0 A (from a V-I table that plots as a straight line through the origin). Find the resistance. Solution: The slope $V/I$ is constant. Using one clean pair, $R = V/I = 3.4/1.0 = 3.4\ \Omega$. (Any pair on the line gives about the same value, roughly 3.4 Ω.)