Earth wire is a safety measure that prevents short circuits and shock. When a live wire touches the metallic case of an electric gadget, the electric current flows from the casing of the appliance to the earth through the copper wire. As the earth offers very low or almost no resistance to the flowRead more

Earth wire is a safety measure that prevents short circuits and shock. When a live wire touches the metallic case of an electric gadget, the electric current flows from the casing of the appliance to the earth through the copper wire. As the earth offers very low or almost no resistance to the flow of current, so large current passes through the copper wire instead of the human body. Due to this large current heat is produced in the circuit and hence the fuse in the circuit melts. So, the circuit is switched off automatically and the electric appliance is saved from burning and no electric shock to the human body.

Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω Potential difference, V = 9 V Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A ∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A
∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2= 3ohm which is also not desired. Hence, we should eitRead more

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be

6/2= 3ohm

which is also not desired. Hence, we should either connect the two resistors in series or parallel.

(i) Two resistors in parallel

Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be

1/(1/6 + 1/6) = 3ohm

The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.

(ii) Two resistors in series

Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be

Here, current, I = 5 A, voltage, V = 220 V ∴ Maxium power, P = I x V = 5 x 220 = 1100W Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110 ∴ 110 lamps can be connected in parallel.

Here, current, I = 5 A, voltage, V = 220 V
∴ Maxium power, P = I x V = 5 x 220 = 1100W
Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110
∴ 110 lamps can be connected in parallel.

Supply voltage, V = 220 V Resistance of one coil, R = 24 Ω (i) Coils are used separately According to Ohm’s law, V = I1R1 Where, I1 is the current flowing through the coil 𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴 Therefore, 9.16 A current will flow through the coil when used separately. (ii) Coils are connectRead more

Supply voltage, V = 220 V

Resistance of one coil, R = 24 Ω

(i) Coils are used separately

According to Ohm’s law,

V = I_{1}R_{1}

Where,

I_{1 }is the current flowing through the coil

𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴

Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω

According to Ohm’s law,

V = I_{2}R_{2}

Where,

I_{2} is the current flowing through the series circuit

𝐼=𝑉𝑅=220/48=55/12=4.58 𝐴

Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel

Total resistance, R_{3} is given as

1/𝑅=1/24+1/24=2/24=1/12 ⟹𝑅=12 Ω

According to Ohm’s law,

V = I_{3}R_{3}

Where,

I_{3 }is the current flowing through the circuit

𝐼=𝑉𝑅=22012=553=18.33 𝐴

Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

## What is the function of an earth wire? Why is it necessary to earth metallic appliances?

Earth wire is a safety measure that prevents short circuits and shock. When a live wire touches the metallic case of an electric gadget, the electric current flows from the casing of the appliance to the earth through the copper wire. As the earth offers very low or almost no resistance to the flowRead more

Earth wire is a safety measure that prevents short circuits and shock. When a live wire touches the metallic case of an electric gadget, the electric current flows from the casing of the appliance to the earth through the copper wire. As the earth offers very low or almost no resistance to the flow of current, so large current passes through the copper wire instead of the human body. Due to this large current heat is produced in the circuit and hence the fuse in the circuit melts. So, the circuit is switched off automatically and the electric appliance is saved from burning and no electric shock to the human body.

See less## A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R ‘, then the ratio R/R ‘ is –

(d) 25

(d) 25

See less## An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(d) 25 W

(d) 25 W

See less## Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(c) 1 : 4

(c) 1 : 4

See less## How is a voltmeter connected in the circuit to measure the potential difference between two points?

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

See less## When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Here, V = 12 V and I = 2.5 mA = 2.5 x 10-3 A ∴ Resistance, R = V/I = 12V/(2.5× 103A) = 4,800 Ω = 4.8 x 10-3 Ω

Here, V = 12 V and I = 2.5 mA = 2.5 x 10

See less^{-3}A∴ Resistance, R = V/I = 12V/(2.5× 10

^{3}A) = 4,800 Ω = 4.8 x 10^{-3}Ω## A battery of 9 V is connected in series with resistors of 0.2 ohm, 0.3 ohm, 0.4 ohm , 0.5 ohm and 12 ohm, respectively. How much current would flow through the 12 ohm resistor?

Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω Potential difference, V = 9 V Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A ∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω

See lessPotential difference, V = 9 V

Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A

∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

## Show how you would connect three resistors, each of resistance 6 W, so that the combination has a resistance of (i) 9 ohm, (ii) 4 ohm.

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2= 3ohm which is also not desired. Hence, we should eitRead more

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be

6/2= 3ohm

which is also not desired. Hence, we should either connect the two resistors in series or parallel.

(i) Two resistors in parallel

Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be

1/(1/6 + 1/6) = 3ohm

The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.

(ii) Two resistors in series

Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be

1/(1/12 + 1/6) = 4ohm

Therefore, the total resistance is 4 Ω.

See less## Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Here, current, I = 5 A, voltage, V = 220 V ∴ Maxium power, P = I x V = 5 x 220 = 1100W Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110 ∴ 110 lamps can be connected in parallel.

Here, current, I = 5 A, voltage, V = 220 V

See less∴ Maxium power, P = I x V = 5 x 220 = 1100W

Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110

∴ 110 lamps can be connected in parallel.

## A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 ohm resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Supply voltage, V = 220 V Resistance of one coil, R = 24 Ω (i) Coils are used separately According to Ohm’s law, V = I1R1 Where, I1 is the current flowing through the coil 𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴 Therefore, 9.16 A current will flow through the coil when used separately. (ii) Coils are connectRead more

Supply voltage,

V= 220 VResistance of one coil,

R =24 Ω(i) Coils are used separately

According to Ohm’s law,

V = I

_{1}R_{1}Where,

I

_{1 }is the current flowing through the coil𝐼=𝑉𝑅=220/24=55/6=9.16 𝐴

Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, R2 = 24 Ω + 24 Ω = 48 Ω

According to Ohm’s law,

V = I

_{2}R_{2}Where,

I

_{2}is the current flowing through the series circuit𝐼=𝑉𝑅=220/48=55/12=4.58 𝐴

Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel

Total resistance, R

_{3}is given as1/𝑅=1/24+1/24=2/24=1/12 ⟹𝑅=12 Ω

According to Ohm’s law,

V = I

_{3}R_{3}Where,

I

_{3 }is the current flowing through the circuit𝐼=𝑉𝑅=22012=553=18.33 𝐴

Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

See less