Physics, asked by MioAmore, 7 months ago

How many electric bulbs each rated as 100 W at 220 V are required
to be joined in parallel with the supply voltage of 220 V to draw 5 A
current from the service mains ?​

Answers

Answered by akbarhussain26
1

Answer:

Question: 12. 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?

Answer:

Given, power = 10W

Potential difference, V = 220 V

Electric current, I = 5A

Here to calculate total number of bulbs, total resistance will be calculated.

We know that,

electricity-exercise solution22

Let, number of bulb connected into parallel = m

Let, total effective resistance, when m bulbs are connected in parallel = R1

electricity-exercise solution23

Therefore, total number of bulbs = 110 Answer

Question: 13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

Case : 1: When coil is used separately, then

Resistance, R = 24 Ω and V = 220V

electricity-exercise solution24

Thus, in this case electric current, I = 9.16 A

Case : 2: When the two resistors are connected in series.

Total effective resistance in the series when coils are used in series= 24 Ω + 24Ω = 48 Ω

electricity-exercise solution25

Thus, in this case electric current, I = 4.58 A

Case : 3: When the two resistors are connected in parallel,

Let the total effective resistance = R

electricity-exercise solution26

electricity-exercise solution27

(a) When coil is used separately the electric current (I) = 9.16A

(b) When coils are used in series the electric current (I)= 4.58 A

(c) When coils are used in parallel the electric current (I)= 18.33 A

Question: 14. Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors

The total effective resitance, R = 1 Ω + 2 Ω = 3 Ω

electricity-exercise solution28

Since, electric current remains the same when resistors are connected in series.

Therefore, electric current through the resistor of 2Ω = 2A

electricity-exercise solution29

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors

Since, voltage across the circuit remains same if resistors are connected in parallel.

electricity-exercise solution30

Power used by resistance of 2 Ω in both the case = 8 W

Similar questions