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Solution : 

Given :- 1 amp current is flowing through 5ohm resistor. 

We know that in case of parallel connection, the p.d. across each resistor is same and is equal to the voltage applied. 

Therefore, applied voltage, V = IR = 1 x 5 = 5V 

So, 

Current through 4 ohm resistor = V/R = 5/4 = 1.25 A 

Current through 10 ohm resistor = V/R = 5/10 = 0.5 A

Wire Y, because of R ∝ l

Resistance of a conductor is directly proportional to the length of the conductor, whose area of cross-section is the same.

In series because of R =R₁ + R₂ + R₃  

=5+10+30 =45 Ω

True Human body is a conductor of electricity.

True If the charge is not in motion, we call it static electricity.

1. The cell can supply electricity in a circuit for a period of time. 

2. After that, chemicals present in it get exhausted and it cannot produce electricity.

True The number of electrons and protons in an atom are same.

1. The ordinary bulb we use gives us light and also heat. 

2. This is not desirable. 

3. This results in the wastage of electricity. 

4. Hence, we should replace an ordinary bulb with a LED bulb. 

5. LED bulbs consume less electricity than ordinary bulb.

Power consumed by

5 bulbs = 5 × 100 = 500 W

2 fans = 2 × 60= 120 W

2 A.C = 2 × 1.5 x 1000 = 3000 W

(a) Total power consumed per day = 3620 W

(b) Total power consumed in 30 days = 3620/1000 × 30 = 108.6 kw

(c) Electric energy is used for 5 h per day.

Total electrical energy consumed in 30 days.

(d) Cost of electricity = P × t = 108.6 × 5 = 543 kWh

= 543 × 6.25 = Rs 3393.75

1. Now-a-day every electrical device is marked with a power saving guide which contains star symbols. 

2. The number of stars on them indicates the energy efficiency of that device. 

3. The electrical appliance with more stars consumes less electricity. 

4. Hence, it is advised to select electrical appliances with more stars.

Electric energy consumed per day = 12 kWh = 12 units

Electric energy consumed in 30 days = 12 × 30 = 360

units Cost to be paid in 1 month = 360 × 6.25 = Rs 2250

As R= V/I, clearly, the slope of V-I graph gives I resistance R. As graph B has a greater slope than A, so graph B represents a series combination (higher resistance) and graph A represents parallel combination (lower resistance).

A battery is used to maintain the potential difference across a conductor.

Electrical resistance, R=V/I

The resistance of a conductor is said to be one ohm if the potential difference applied across its ends is 1 volt and a current, of 1 A flows through it. It SI unit is ohm (Ω)

1 Ohm= 1 volt/1 ampere.

I = q/t

t = 20 x 60 = 1200 seconds,

I = 200 mA = 200 X 10^-3 A

I = 200 mA  = 200 X 10^-3A

Charge passing = q = lt = 200 x 10^-3 x 1200 = 240 C

(a) When 1 A current flows across the wire in 1 second, the charge transfer across its ends is said to be 1 coulomb. 

(b) The relationship between the electric current I, the charge q and time t is

Correct Answer is:  (b, d)

1. Cells are connected in parallel. 

2. The bulb will glow as usual if one of the cells is disconnected.

If a circuit is closed that means electrons can easily flow through the circuit and thus electric current will flow through it. 

If a circuit is open that means the electrons will not be able to flow through the circuit which means that there will be no current flowing through it.

Voltmeter is a high resistive device that is used to measure the voltage across a component. It is always connected in parallel to the component whose voltage is to be measured. It is highly resistive so that the current cannot pass through it and thus the current will pass through the component connected in parallel to it so as to measure the accurate voltage drop across the component.

Ammeter is a device that is connected in series so as to measure the current through the circuit. It has a very low resistance so that maximum current passes through it and thus the correct amount of current through the circuit can be measured.

Correct Answer is: (c) 32 μV 

Potentials of the two spheres ±  e/C.

∴ potential difference = 2e/C = 2e/4πε₀r = 9 x 10⁹ 2 x 1.6 x 10^-19 / 9 x 10^-5 V

= 3.2 x 10^-5 V = 32 μV.

Correct Answer is: (d) 2Q/3

The capacitances are C₁ = 4πε₀R and C₂ = 4πε₀(2R) = 2C₁.

The common potential = V = Q/ C₁ + C₂.

The charge flowing = charge on C₂

= C₂ V = Q C₂/C₁ + C₂ = Q 2C₁/ C₁ + 2C₁ = 2/3 Q.

Correct Answer is: (c, d)

The capacitances of the two are C₁ = 4πε₀R and C₂ = 4πε₀(2R).

 The initial energy, E_i = Q²/2C₁.

The final energy, E_f = Q²/2C₂.

The heat produced = E_i - E_f = [(1/4πε₀R) - (1/2 x πε₀R)]

= k Q²/2R [1 - 1/2] = KQ²/4R.

60 W bulb has a higher resistance than the resistance across 100 W bulb since the power developed is P = \(\frac{V^2}{R}\) . Potential difference across a bulb will be proportional to resistance. Hence potential difference across 60 W bulb is higher than that across 100 W bulb.

• Property of the conductor which resists the flow of electric current is called resistivity. 
• Resistance for a particular material is unique. 
• Resistance is directly proportional to length of conductor and inversely proportional to current flow. 
• When length is doubled resistance becomes double and current flow reduces to half. 
• This is the reason for the decrease in ammeter reading.

The SI unit of electrical conductivity is Siemens / metres (s/m).

Correct answer is (c) 250 mA

I = 5A 

IA = 106 µA 

5A = 5 × 106 µA 

5A = 50,00,000 µA

1. (b) 

2. (d) 

3. (a) 

4. (c)

1. (d) 

2. (c) 

3. (a) 

4. (b)

Wire is used to connect electrical component to one another.

a. Both A and R are correct and R is the correct explanation for A.