1 kW , 220 V electric heater is to be used with 220 V D . C . supply. (a) What is the current in the heater? (b) What is its resistance? (c) What is the power dissipated in the heater? (d) How much heat in calories is produced per second? (e) How many grams of water at 100 ° C will be converted per minute into steam at 100 ° C ?

Given voltage to the heater, V = 220 V , Power rating of the heater, P = 1 kW = 1000 W , Now, (a) We know that power generated by a resistive load with certain current, I and voltage supply, V is given as, P = V I , So, P = V × I ⇒ 1000 = 220 × I ⇒ I = 1000 220 = 4 . 55 A . Hence, the current in the heater is 4 . 55 A . (b) Power generated in the heater can be given as, P = V 2 R ( R is the resistance of the heater ) ⇒ 1000 = 220 × 220 R ⇒ R = 48400 1000 = 48 . 4 Ω . Hence, the resistance of the heater is 48 . 4 Ω . (c) As given power rating of the heater is 1000 W so this is the power dissipated in the heater which can also be given as, P = V 2 R = 220 × 220 48 . 4 = 1000 W . (d) Heat, H , produced per second of the heater, i.e. H = V 2 R × t = 220 × 220 48 . 4 × 1 = 1000 J . Since 1 J = 1 4 . 2 calories , So, 1000 J = 1 4 . 2 × 1000 = 238 . 1 calories . So, heat in calories produced per sec is 238 . 1 cal s - 1 . (e) Heat produced per second is 238 . 1 calories s - 1 (as obtained above) so the same heat will be converted to steam and if taken per minute then, we get, 238 . 1 × 60 = 14286 calories min - 1 . Since, the water at 100 C o is converted to steam at 100 C o , and we know that during phase change the heat energy is released or absorbed is given as, Q = m L , Here Q = 14286 calories , Latent heat of water L = 540 cal g - 1 m is the mass of water. So, m = Q L ⇒ m = 14286 540 = 26 . 5 g . Hence, mass of water at 100 C o converted per minute into steam at 100 C o is 26 . 5 g .

In following circuit potential at point ‘ A ’ is zero then determine (a) Potential at each point. (b) Potential difference across each resistance. (c) Identify the batteries which act as a source. (d) Current in each battery. (e) Which resistance consumes maximum power? (f) Which battery consume or gives maximum power?

(a) Given that potential at A = 0 and potential difference between E and A is 10 V which is given as, V E - V A = 10 ⇒ V E = 10 V It is clear from the given circuit diagram that points E , F , G , H are connected to one another, so the potential at these points will be equal i.e; 10 V . Similarly, points A , B , C , D are connected to one another, so the potential at these points will be same i.e; 0 V . As 10 V supply is in series with 5 V supply between points F and B , So, the potential at I = 10 + 5 = 15 V . Now, potential at J = 15 V . Similarly, potential at K will be the sum of potentials at H and the 5 V source (as connected in series) i.e; V K = 10 + 5 = 15 V . (b) Potential difference across resistor connected between points F and B is V I - V B = 15 - 0 = 15 V , Potential difference across resistor connected between points G and C is V J - V G = 15 - 10 = 5 V , Potential difference across resistor connected between points H and D is V K - V D = 15 - 0 = 15 V . (c) From the given circuit diagram we can see that the 10 V source battery is in series with the two 5 V batteries which are in parallel also the 15 V battery is in series with the other two 5 V batteries, So, all the batteries act as source. (d) Current in the battery between points F and B is = 15 1 = 15 A , Current in the battery between points G and C is = 5 2 = 2 . 5 A , Current in the battery between points H and D is = 15 3 = 5 A , Now, for current (say I ) in the battery of 10 V , lets reduce all the other batteries into current source as shown in the diagram below and applying K C L at point O then, we get, I + 2 . 5 - 15 - 5 = 0 ⇒ I = 20 - 2 . 5 = 17 . 5 A . (e) Power in each resistor is given as, For 1 Ω resistor, P = V 2 R = 15 2 1 = 225 W , For 2 Ω resistor, P = V 2 R = 5 2 2 = 12 . 5 W , For 3 Ω resistor, P = V 2 R = 15 2 3 = 225 3 = 75 W . Hence, we can see 1 Ω resistor consumes maximum power. (f) Power in each battery is given as, For 10 V battery, P = V I = 10 × 17 . 5 = 175 W , For 5 V battery, P = V I = 5 × 15 = 75 W , For 15 V battery, P = V I = 15 × 2 . 5 = 37 . 5 W , For last 5 V battery at the extreme right side of the circuit, P = 5 × 5 = 25 W . So, it is clear that 10 V battery uses maximum power.

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$(i)$ A potential difference of $200 volt$ is applied to a coil at a temperature of $15 ° C$ and the current is $10 A$ . What will be the temperature of the coil when the current has fallen to $9 A$ , the applied voltage being the same as before? Temperature coefficient of resistance $(α) = \frac{1}{234} ° C^{- 1}$ .

$(ii)$ A platinum wire has a resistance of $10 ohm at 0 ° C$ and $20 ohm at 273 ° C$ . Find the value of the temperature coefficient of resistance.

Important Questions on Current Electricity

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 22

The current-voltage graphs for a given metallic wire at two different temperatures $T_{1}$ and $T_{2}$ are shown in the figure. Which one is higher, $T_{1}$ or $T_{2}$

Question Image

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 23

If a copper wire is stretched to make it

0.1 %

longer, what is the percentage change in its resistance?

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 24

A rectangular carbon block has dimensions

1.0 cm \times 1.0 cm \times 50 cm

(i)

What is the resistance measured between the two square ends?

(ii)

Between two opposing rectangular faces?
The resistivity of carbon at

20 ° C is 3.5 \times 10^{- 5} Ω m

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 25

In following diagram boxes may contain resistor or battery or any other element.

Question Image

Then determine in each case

(a) $E . m . f .$ of battery.

(b) Battery is acting as a source or load.

(d) Power input to the battery or output by the battery.

(e) The rate at which heat is generated inside the battery.

(f) The rate at which the chemical energy of the cell is consumed or increased.

(g) Potential difference across the box

(h) Electric power output across the box.

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 26

A resistor with a current of

3 A

through it converts

500 J

of electrical energy to heat energy in

12 s

. What is the voltage across the resistor?

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 27

(a) A car has a fresh storage battery of emf $12 V$ and internal resistance $5.0 \times 10^{- 2} Ω$ . If the starter draws a current of $90 A$ , what is the terminal voltage of the battery when the starter is on?

(b) After long use, the internal resistance of the storage battery increases to $500 Ω$ . What maximum current can be drawn from the battery? Assume the emf of the battery remain unchanged.

(c) If the discharged battery is charged by an external emf source, is the terminal voltage of the battery during charging greater or less than its emf $12 V$ ?

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 28

$1 kW, 220 V$ electric heater is to be used with $220 V D . C .$ supply.

(a) What is the current in the heater?

(b) What is its resistance?

(d) How much heat in calories is produced per second?

(e) How many grams of water at $100 ° C$ will be converted per minute into steam at $100 ° C$ ?

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 1>Q 29

In following circuit potential at point $‘ A ’$ is zero then determine

Question Image

(a) Potential at each point.

(b) Potential difference across each resistance.

(d) Current in each battery.

(e) Which resistance consumes maximum power?

(f) Which battery consume or gives maximum power?

Important Questions on Current Electricity

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