1. Rate of flow of electric charge is called electric
current. SI unit: Ampere (A).
2. A continuous and closed path of an electric current.
(Diagram: Draw battery, switch, bulb, ammeter in series).
[Diagram: Simple Circuit]
3. $Q = ne$ $\Rightarrow n = Q/e$ $= 1 / (1.6 \times
10^{-19})$ $= 6.25 \times 10^{18}$ electrons.
4. $Q = It$. $I=0.5$A, $t=10 \text{ min} = 600$s.
$Q
= 0.5 \times 600 = 300$ C.
5. Conventional current flows opposite to the direction of
flow of electrons.
6. To create a potential difference which makes the charge
flow.
7. One Ampere is the current flowing when 1 Coulomb of
charge flows through a cross-section in 1 second.
8. Ammeter. Connected in series.
9. Potential difference across the conductor.
10. $n=10^{20}$, $t=1$s.
$Q=ne$ $= 10^{20} \times 1.6
\times 10^{-19}$ $= 16$ C.
$I = Q/t = 16/1 = 16$ A.
11. Open: Key is off, current doesn't flow. Closed: Key is
on, current flows.
12. Scalar. It has direction but follows algebraic sum
laws, not vector addition.
13. Free electrons.
14. Charge exists in discrete packets; total charge is
integral multiple of charge of electron ($Q=ne$).
15. $I = Q/t = 60 / (2 \times 60) = 60/120 = 0.5$ A.
16. Work done to move unit positive charge from one point
to another. Device: Voltmeter.
17. At constant temperature, current is directly
proportional to potential difference. $V=IR$.
[Diagram: Ohm's Law Verification]
18. $W = VQ = 15 \times 3 = 45$ J.
19. 1 Volt is 1 Joule of work done to move 1 Coulomb
charge. (1 J/C).
20. Nuclear forces hold electrons. Conductors have free
electrons that move easily.
21. Straight line passing through origin. Slope represents
Resistance ($R$).
22. $I = V/R = 220 / 50 = 4.4$ A.
23. Curved line. Example: Diode or Filament bulb.
24. According to Ohm's law ($I \propto V$), if V doubles,
Current also doubles.
25. Because it measures PD across two points and has high
resistance so it doesn't draw current.
26. Infinite resistance.
27. $V_{AB} = V_A - V_B = 100 - (-10) = 110$ V. $W = QV =
5 \times 110 = 550$ J.
28. No. Fails for Diodes, Transistors, Electrolytes.
29. Collision of drifting electrons with atoms/ions
generates heat.
30. $V=12$V, $I=2.5$mA = $0.0025$A. $R = V/I = 12 / 0.0025
= 4800 \Omega = 4.8 \text{ k}\Omega$.
31. Length, Area, Nature of material, Temperature.
32. Alloys have higher resistivity and do not oxidize
(burn) easily potential difference high temperatures.
33. Volume constant. If $L \to 2L$, then $A \to A/2$.
$R_{new} \propto (2L)/(A/2)$ $= 4(L/A) = 4R$. Resistance becomes 4 times.
34. Series: $V = V_1+V_2+V_3$
$\Rightarrow IR_s =
IR_1+IR_2+IR_3$
$\Rightarrow R_s = R_1+R_2+R_3$.
35. Parallel: $I = I_1+I_2+I_3$
$\Rightarrow V/R_p =
V/R_1+V/R_2+V/R_3$
$\Rightarrow 1/R_p = 1/R_1+1/R_2+1/R_3$.
36. (i) If one device fails, all stop. (ii) All get
different voltages if resistances vary (voltage division).
37. Same resistors in parallel: $R_{eq} = R/n = 10/2 = 5
\Omega$.
38. Cut into 5 parts $\to$ each part $R/5$. Parallel:
$1/R' = 5/(R/5) = 25/R \Rightarrow R' = R/25$. Ratio $R/R' = 25$.
39. $R = \rho L/A \Rightarrow \rho = RA/L$. $R=2$, $L=1$.
$A = \pi (d/2)^2$ $= 3.14 \times (2 \times 10^{-4})^2$ $= 3.14 \times 4 \times 10^{-8}$ $= 1.256 \times
10^{-7} m^2$.
$\rho = 2 \times 1.256 \times 10^{-7} / 1$ $= 2.5 \times 10^{-7} \Omega$ m.
40. Resistance is property of the object (depends on
dimensions). Resistivity is property of material (independent of dimensions).
41. $P = V^2/R \Rightarrow R \propto 1/P$. Lower power
means higher resistance. 60W bulb has higher resistance.
42. $1/R = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1$. $R_{eq}
= 1 \Omega$.
43. Circuit breaks, current stops, all stop working.
44. (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 ($12||6 \Rightarrow
(12\times6)/(12+6) = 72/18 = 4$).
45. It means resistance of a copper wire of length 1m and
area $1m^2$ is $1.62 \times 10^{-8} \Omega$.
46. Heat produced $H = I^2 R t$. (Directly proportional to
square of current, resistance, and time).
47. Cord connects (Copper - low R, negligible heat).
Element heats (Nichrome - high R, high heat).
48. $H = VQ$ (or VIt). given Q=96000, t=1hr (irrelevant
here if we use VQ), V=50. $H = 50 \times 96000 = 4,800,000$ J = $4.8 \times 10^6$ J.
49. Max rate ($P=840$): $I = 840/220$ $= 3.82$ A,
$R
= 220/3.82$ $= 57.6 \Omega$.
Min rate ($P=360$): $I = 360/220$ $= 1.64$ A,
$R = 220/1.64$ $=
134.15 \Omega$.
50. Electric Power.
51. $P = VI = 220 \times 5 = 1100$ W. Energy $E = P \times
t = 1.1 \text{ kW} \times 2 \text{ h} = 2.2$ kWh.
52. kWh. $1 \text{ kWh} = 3.6 \times 10^6$ J.
53. Low resistivity minimizes energy loss as heat.
54. TV: $250 \times 1 = 250$ Wh. Toaster: $1200 \times
(10/60) = 1200/6 = 200$ Wh. TV uses more.
55. Works on heating effect. Melts if current exceeds
rating. Rated in Amperes (e.g., 5A, 10A).
56. $R = V^2/P = 220^2 / 100 = 48400 / 100 = 484 \Omega$.
57. Series ($R_s = 2R$), Parallel ($R_p = R/2$). Same V.
$H = V^2 t / R$. Ratio $H_s/H_p$ $= (V^2 t / 2R) / (V^2 t / (R/2))$ $= (1/2R) \times (R/2)$ $= 1/4
\Rightarrow 1:4$.
58. By Joule's Law derivation, work done against
resistance is converted to heat, and power $P \propto I^2$.
59. Argon or Nitrogen (Chemicaly inactive). To prolong
filament life (prevent oxidation).
60. $E \text{ per day} = 400 \times 8 = 3200 \text{ Wh} =
3.2 \text{ kWh}$. For 30 days: $3.2 \times 30 = 96$ kWh. Cost: $96 \times 3 = \text{Rs } 288$.
