1. Battery (3 cells series = 6V), R1=5, R2=8, R3=12, Key.
All in series.
[Schematic: 6V Battery - Key - 5Ω - 8Ω - 12Ω]
2. $R_{eq} = 5+8+12=25 \Omega$. $I = V/R = 6/25 = 0.24$ A
(Ammeter).
Voltage across 12$\Omega$: $V = I \times R = 0.24 \times 12 = 2.88$ V (Voltmeter).
3. (i) 9 $\Omega$: Two in parallel ($6||6=3$) + One in
series ($3+6=9$).
(ii) 4 $\Omega$: Two in series ($6+6=12$) parallel with third ($12||6 \Rightarrow
72/18 = 4$).
4. Diagram showing 2, 3, 6 ohm resistors connected across
common points A and B. $1/R = 1/2+1/3+1/6 = 1 \Omega$.
[Diagram: Parallel resistors]
5. Circle wire of R divided into two semicircles of R/2
each. Diameter points connect these two in parallel. $R_{eq} = (R/2 || R/2) = (R/2)/2 = R/4$.
6. (i) Max: Series ($nR$). (ii) Min: Parallel ($R/n$).
Ratio: $nR / (R/n) = n^2$.
7. Parallel part: $3||6$ $\Rightarrow (3 \times 6)/(3+6) =
2 \Omega$.
Total $R = 2 + 4 = 6 \Omega$.
8. $I = V_{bat} / R_{tot} = 12 / 6 = 2$ A.
9. Current is 2A (series). $V_4 = I \times R = 2 \times 4
= 8$ V.
10. PD across parallel combo = $V_{bat} - V_4 = 12 - 8 =
4$ V. Current in 6 $\Omega$ = $V/R = 4/6 = 0.67$ A.
11. Ammeter in series with loop. Voltmeter in parallel
across the specific resistor.
[Diagram: Ammeter Series, Voltmeter Parallel]
12. In series, there is only one path. If one fuse blows,
path is broken (open circuit).
13. Parallel ensures: (1) If one appliance fails, others
work. (2) Each appliance gets full mains voltage (220V). (3) Total resistance decreases, current potential
is sufficient.
14. Diagram: Two bulbs parallel to 220V source. Switches
for each.
[Diagram: Parallel Lamps]
15. Safety device to prevent excess current. Placed in
series with Live wire. Symbol: Rectangle with line through it or sine wave inside.
16. $I = V/R = 250 / 500 = 0.5$ A.
17. $V = IR = 2 \times 20 = 40$ V.
18. $R = V/I = 0.8 / 0.2 = 4 \Omega$.
19. $\rho = RA/L$ $= (2 \times 10^{-2} \times 1.7 \times
10^{-6}) / 2$ $= 1.7 \times 10^{-8} \Omega$ m.
20. Volume const. $L \to 2L$, $A \to A/2$. $R_{new} =
\rho(2L)/(A/2) = 4(\rho L/A) = 4R$. $4 \times 20 = 80 \Omega$.
21. (i) Min Current: Series ($15 \Omega$). $I = 6/15 =
0.4$ A. (ii) Max Current: Parallel ($5||10 = 3.33 \Omega$). $I = 6/3.33 = 1.8$ A.
22. (i) Separate: $I = 220/24 = 9.17$ A. (ii) Series ($48
\Omega$): $I = 220/48 = 4.58$ A. (iii) Parallel ($12 \Omega$): $I = 220/12 = 18.33$ A.
23. (i) Series: $R_{eq}=3, I=2A$.
Power in 2$\Omega$
= $I^2R$ $= 4 \times 2 = 8$ W.
(ii) Parallel: Voltage is 4V across each.
Power in 2$\Omega$ =
$V^2/R$ $= 16/2 = 8$ W. Result: Equal Power (8W).
24. Total Power $P = 100+60 = 160$ W. $I = P/V = 160/220 =
0.727$ A.
25. Energy TV = $250 \times 1 = 250$ Wh. Toaster = $1200
\times (10/60) = 200$ Wh. TV uses more.
26. Rate of heat = Power. $P = I^2 R = 15^2 \times 8 = 225
\times 8 = 1800$ W. (Time is irrelevant for 'rate').
27. $H = V^2 t / R$. $t=1$.
$100 = V^2 / 4$
$\Rightarrow V^2 = 400$ $\Rightarrow V = 20$ V.
28. $P = VI = 220 \times 0.50 = 110$ W.
29. $P=0.4$ kW. Time = $8 \times 30 = 240$ h. Energy =
$0.4 \times 240 = 96$ kWh. Cost = $96 \times 3 = \text{Rs } 288$.
30. $E = P \times t$ $= 500 \text{ W}$ $\times (4 \times
3600) \text{ s}$ $= 500 \times 14400$ $= 7,200,000$ J.
31. (i) $P = VI$ $= 2.5 \times 0.75 = 1.875$ W.
(ii)
$R = V/I$ $= 2.5 / 0.75 = 3.33 \Omega$.
(iii) $E = P \times t$ $= 1.875 \times 4 = 7.5$ Wh.
32. $H = VQ$ $= 50 \times 96000$ $= 4,800,000$ J $= 4.8
\times 10^6$ J.
33. $E = VIt$ $= 12 \times 4 \times (10 \times 60)$ $= 48
\times 600$ $= 28,800$ J.
34. Rated $I = P/V = 60/250 = 0.24$ A.
$R = V^2/P =
250^2/60 = 1041.6 \Omega$.
At 200V: $P' = V'^2/R = 200^2 / 1041.6$ $= 40000 / 1041.6 = 38.4$ W. Power
drops.
35. Formula $P = V^2/R \Rightarrow$ Lower Wattage has
higher Resistance. Bulb B (10W) has higher R. In series, $P = I^2 R$. Higher R consumes more power. So, 10W
bulb glows brighter.
36. (1) Very high melting point ($3380^{\circ}$C), doesn't
melt at white heat. (2) High resistivity, retains heat.
37. Alloys have higher resistivity (produce more heat) and
do not oxidise (burn) easily at high temperatures compared to pure metals.
38. Voltage division makes appliances dim; if one fails,
circuit breaks; total resistance increases too much.
39. $R \propto 1/A$. More area = more path for electrons =
less collisions = less resistance.
40. Rubber is an insulator; prevents current from flowing
through the electrician's body to earth.
41. Extremely low resistivity to minimize power loss
($I^2R$) during transmission.
42. Temp increase $\to$ atoms vibrate more amplitude $\to$
more collisions with drifting electrons $\to$ higher resistance.
43. To melt quickly when current exceeds the safe limit,
breaking the circuit.
44. So it doesn't change the current of the circuit when
placed in series.
45. So it draws negligible current when placed in
parallel.
46. Provides a low-resistance path to ground for leakage
current, protecting the user from shock.
47. Alloy resistivity is weak function of temperature
(almost constant). Pure metals increase significantly.
48. Independent operation and full voltage to each bulb.
49. Drawing more current than the circuit rating (e.g.,
connecting too many high-power appliances). Causes heating/fire.
50. Live and Neutral wires touch directly ($R \approx 0$).
Current becomes infinite. Fuse melts due to high heat ($I^2Rt$) and breaks circuit.
