Electricity is used to operate many electrical devices such as fan, refrigerator, bulb and many more. Electric energy can be changed to different forms of energy like heat, light or mechanical energy, in order to be useful. Electricity is produced by the movement of electrons through the wire. For the electric current to flow the circuit must be closed, in other words there must be an uninterrupted path from the power source through the circuit and then back to the source of power. The branch of physics that deals with charges in motion is termed as current electricity.

Any motion of charges from one section to another is current. When two bodies at different potentials are linked with a wire, free electrons stream from one body to the other body, until both the objects reach the same potential, after which the current stops flowing. Until a potential difference is present throughout a conductor, current runs. *Difference in electric potential is called voltage and is measured in volts*.

The force that opposes the flow of electric current is called **resistance** and is measured in *ohms*. The electric current lost due to resistance can be transformed to heat energy. The heat energy produced is used in an electric stove.

Download Current Electricity Previous Year Solved Questions PDF

## JEE Main Previous Year Solved Questions on Current Electricity

**Q1: Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm ^{2} is v. If the electron density of copper is 9 × 10^{28}/m^{3} the value of v in mm/s is close to (Take charge of electron to be = 1.6 × 10^{–19} C) **

(a) 3

(b) 0.2

(c) 2

(d) 0.02

**Solution**

As I = neAvd = neAv

v = I/neA = 1.5/(9 x 10^{28} x 1.6 x 10^{-19} x 5 x 10^{-6})

v = 0.02 x 10^{-3} m/s = 0.02 mm/s

**Answer: (d) 0.02 **

**Q2: Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be **

(a) 240 W

(b)120 W

(c) 60 W

(d) 30 W

**Solution**

The power consumed when two resistance are in series combination is

V^{2}/2R = 60 W ⇒ V^{2}/R=120 W

When the two resistance are connected in parallel combination, power consumed is

2V^{2}/R = 120(2)= 240 W

**Answer: (a) 240 W **

**Q3: A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is **

(a) 11 × 10^{–4} W

(b) 11 × 10^{–5 }W

(c) 11 × 10^{5} W

(d) 11 × 10^{–3} W

**Solution**

Case (1)

As I^{2}R = P

R = P/I^{2}

R = (4.4)/(2 x 10^{-3})^{2}= 1.1 x 10^{6} Ω

Case (2)

P = V^{2}/R = (1.1)2/(1.1 x 10^{6}) = 11 x 10^{-5} W

**Answer: (b) 11 × 10 ^{–5 }W**

**Q4: An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 . The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire is **

(a) 490

(b) 495

(c) 395

(d) 480

**Solution**

Let I be the current in the circuit. 4 = (5 + R) I——(1)

According to given condition,

5 x 10^{-3} = (10/100)(5)(I)

I = 10^{–2} A …(2)

Using (1) and (2), 5 + R = 400 R = 395

**Answer: (c) 395**

**Q5: A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when **

(a) R = 0.001 r

(b) R = r

(c) R = 2r

(d) R = 1000 r

**Solution**

The power delivered to resistance is I^{2}R

i.e., P = [ε^{2}/(R + r)^{2}]R

For the maximum power,dP/dR= 0

⇒ -2R + (R +r) = 0 or R = r

**Answer: (b) R = r **

**Q6: A metal wire of resistance 3 is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be **

(a) (7/2) Ω

(b) (5/2) Ω

(c) (12/5) Ω

(d) (5/3) Ω

**Solution**

R = 3 Ω = ρ(l/A) = ρ(I^{2}/V)

R’ = ρ(I’^{2}/V) ⇒R’ = (2l)^{2}/l^{2} x 3 ⇒R’ =12 Ω

Equivalent resistance, R_{eq} = (10 x 2)/(10+2) = (5/3) Ω

**Answer: (d) (5/3) Ω**

**Q7: On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of the combination is 1 kΩ. How much was the resistance on the left slot before the interchange? **

(a) 990

(b) 505

(c) 550

(d) 910

**Solution**

Let R_{1} (left slot) and R_{2} (right slot) be two resistances in two slots of a meter bridge. Initially l be the balancing length

Then, R_{1}/R_{2} = l/(100 – l)———-(1)

R_{1} + R_{2} = 1000 Ω———-(2)

On interchanging the resistances, balancing length becomes (l – 10), so

( R_{2}/R_{1})= (l – 10)/(110 – l)

Using (1)

(100 – l)/l = (l – 10)/(110 – l)

11000 + l^{2} – 210 l = l^{2} – 10l

200 l = 11000, l = 55 cm

From equa (1) R_{1}/R_{2} = 55/45

Using (2)

R_{1} = (55/45)(1000 – R_{1})

R_{1} + (55/45)R_{1 }= (1000)x(55/45)

100 R_{1} = 1000 x 55

R_{1} = 550 Ω

**Answer: (c)550 **

**Q8: A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be **

(a) Increased 8 times

(b) Unchanged

(c) Doubled

(d) Halved

**Solution**

Rate of heat developed, P = V^{2}/R

For given V, P ∝ 1/R = A/ρl = πr^{2}/ρl

Now, P_{1}/P_{2} = (r_{1}^{2}/r_{2}^{2})(l_{2}/l_{1})

As per question, l_{2} = l_{1}/2 and r_{2} = 2r_{1}

P_{1}/P_{2} = (¼) x (½) = ⅛

P_{2} = 8P_{1}

**Answer: (a) Increased 8 times **

**Q9: A heating element has a resistance of 100 Ω at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and the temperature is 500°C more than room temperature. What is the temperature coefficient of resistance of the heating element? **

(a) 1×10^{– 4} °C^{–1}

(b) 2 × 10^{– 4} °C^{–1 }

(c) 0.5 × 10^{– 4} °C^{–1 }

(d) 5 × 10^{– 4 }°C^{–1 }

**Solution**

Resistance after temperature increases by 500°C,

R_{T} = Voltage applied/Current = 220/2 = 110

Also, R_{T} = R_{0} (1 + αΔT)

110 = 100 (1 + (α x 500))

α = 10/(100 x 500) = 2 x 10^{-4 0}C^{-1}

**Answer: (b) 2 × 10 ^{– 4} °C^{–1 } **

**Q10: A uniform wire of length l and radius r has a resistance of 100 Ω. It is recast into a wire of radius r/2.The resistance of new wire will be **

(a) 400 Ω

(b)100 Ω

(c) 200 Ω

(d)1600 Ω

**Solution**

Resistance of a wire of length l and radius r is given by

R = ρl/A = (ρl/A) x(A/A)

R = (ρl/A^{2}) = (ρV/π^{2}r^{4}) (∵ V = Al)

i.e., R ∝ 1/r4

R_{1}/R_{2} =(r_{2}/r_{1})^{4}

Here, R_{1} = 100 Ω , r_{1} = r, r_{2} = r/2

R_{2}= R_{1}(r_{1}/r_{2})^{4} = 16R_{1} = 16000 Ω

**Answer: (d) 1600 Ω **

**Q11: A 2 W carbon resistor is colour coded with green, black, red and brown, respectively. The maximum current which can be passed through this resistor is **

(a) 20 mA

(b) 0.4 mA

(c) 100 mA

(d) 63 mA

**Solution**

The resistance of the resistor is 50 × 10^{2 }Ω . So, the maximum current that can be passed through it is

**Answer: (a) 20 mA **

**Q12: In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be **

(a) 14 A

(b) 8 A

(c) 10 A

(d) 12 A

**Solution **

Power of 15 bulbs of 40 W = 15 × 40 = 600 W

Power of 5 bulbs of 100 W = 5 × 100 = 500 W

Power of 5 fan of 80 W = 5 × 80 = 400 W

Power of 1 heater of 1 kW = 1000

Total power, P = 600 + 500 + 400 + 1000 = 2500 W

When these combination of bulbs, fans and heater are connected to 220 V mains, current in the main fuse of building is given by

I= P/V = 2500/220 = 11.36 A ≈ 12 A

**Answer: (d) 12 A**

**Q13: If a wire is stretched to make it 0.1% longer, its resistance will **

(a) increase by 0.05%

(b) increase by 0.2%

(c) decrease by 0.2%

(d) decrease by 0.05%

**Solution**

Resistance of wire R = ρl/A……..(1)

On stretching, volume (V) remains constant.

So V = Al or A=V/l

Therefore, R = ρl^{2}/V (Using (1))

Taking logarithm on both sides and differentiating we get,

ΔR/R = 2Δl/l (Since V and ρ are constants)

(ΔR/R)% = (2Δl/l)%

Hence, when wire is stretched by 0.1% its resistance will increase by 0.2%

**Answer: (b) increase by 0.2% **

**Q14: A thermocouple is made from two metals, antimony and bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will **

(a) flow from antimony to bismuth at the cold junction

(b) flow from antimony to bismuth at the hot junction

(c) flow from bismuth to antimony at the cold junction

(d) not flow through the thermocouple.

**Solution:**Antimony-bismuth couple is ABC couple. It means that current flows from A to B at a cold junction

**Answer: (a) flow from antimony to bismuth at the cold junction **

**Q15: The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. The resistance of the wire at 0°C will be **

(a) 3 ohm

(b) 2 ohm

(c) 1 ohm

(d) 4 ohm

**Solution**

R_{t} = R_{0}(1 + αt)

R_{t } is the resistance of wire at t^{0} C

R_{0 }is the resistance of wire a 0^{0} C

α is the temperature coefficient of resistance

R_{50} = R_{0} [1 + α(50)]

And R _{100} = R_{0} [1+ α(100)]

Or R_{50} – R_{0}= R_{0} α(50) ——-(1)

R _{100} – R_{0} = R_{0} α(100) ——-(2)

Dividing (1) by (2)

(5 – R_{0})/((6 – R_{0}) = ½

R_{0} = 4 ohm

**Answer: (d) 4 ohm**