Electricity ncert exercise question solution part 2 Class 10 Science

Electricity

NCERT Solution 2

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:Given, resistance of each of the resistor = 176 Ω
Electric current ( I ) = 5A
Potential difference (V) = 220V
Number of resistors connected in parallel =?

Let total x resistors are connected in parallel, and total effective resistance = R

Therefore, `(1)/(R_t)=x\xx(1)/(176Ω)=(x)/(176Ω)`

Hence, `R_t=(176Ω)/(x)`

We know that `R=V/I`

So, `(176Ω)/(x)=(220V)/(5A)`

Or, `x\xx220V=176Ωxx5A`

Or, `x=(176Ωxx5A)/(220V)=4`

Thus, 4 resistors need to be connected.


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

Answer: Here we have four options to connect the three resistors in different ways.

  1. All the three resistors can be connected in series
  2. All the three can be connected in parallel
  3. Two of the three resistors can be connected in series and one in parallel and
  4. Two of the three resistors can be connected in parallel and one in series

Thus, effective resistance in the case

When all the three resistors are connected in series

Effective total resistance `= 6 Ω + 6 Ω + 6 Ω = 18 Ω`. This is not required

When all the three are connected in parallel

Then, `1/R=(1)/(6Ω)+ (1)/(6Ω)+ (1)/(6Ω)`

`=(1+1+1)/(6)Ω=3/6Ω=1/2Ω`

Thus, effective total resistance R = 2 Ω. This is also not required.


When two of the three resistors are connected in parallel and one in series

When two resistors are connected in parallel

Then, `1/R=(1)/(6Ω)+ (1)/(6Ω)=1/3Ω`

Therefore, R = 3Ω

And third one is connected in series, then total resistance = 3 Ω + 6 Ω = 9 Ω

Two of the three resistors can be connected in parallel and one in series

When two resistors are connected in series, then total resistance `= 6 Ω + 6 Ω = 12 Ω`

And one resistor is connected in series with two in parallel

Then, `1/R=(1)/(12Ω)+ (1)/(6Ω)`

`=(1+2)/(12)Ω=(3)/(12)Ω=1/4Ω`

Thus, R = 4 Ω

Thus, when two resistors are connected in series and one in parallel then total effective resistance = 9 Ω

When two resistors are connected in parallel with one in series then total effective resistance = 4 Ω


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, potential difference (V) = 220 V
Power input (P) = 10W
Allowable electric current (I) = 5A
Number of lamps connected in parallel =?

To calculate this, first of all resistance of each of the lamp is to be calculated.

We know; `P = (V^2)/(R)`
Hence, `10 W = (220 V)^2 ÷ R`
Or, `R = 48400 ÷ 10 W = 4840 Ω`

Let x bulb are to be connected in parallel to have electric current equal to 5 A. Therefore;

So, `1/R=x\xx(1)/(4840Ω)=(x)/(4840Ω)`

Effective resistance `R=(4840)/(x)Ω`

We know that `R=V/I`

Or, `(4840)/(x)Ω=(220V)/(5A)`

Or, `x=(4840xx5)/(220)=110`

Thus, total 110 bulbs need to 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 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer: Given, Potential difference (V) = 220 V
Resistance of each coil = 24 Ω

The current in given three case, i.e. when used separately, when used in parallel, when used in series =?

Case 1: When used separately, then resistance, R = 24 Ω and V = 220V
We know; `I = V/R`
`= 220 V ÷ 24 Ω = 9.16 A`

Case 2: When the two resistors are connected in series.
Total effective resistance = 24 Ω + 24Ω = 48 Ω
Hence, electric current can be calculated as follows:
`I = V/R = 220 V ÷ 58 Ω = 4.58 A`

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

Then, `1/R=(1)/(24Ω)+ (1)/(24Ω)`

`=(1+1)/(24Ω)=(2)/(24Ω)=(1)/(12Ω)`

Therefore, total effective resistance R = 12 Ω
Thus, electric current `I = V/R`
`= 220 V ÷ 12 Ω = 18.33 A`

Thus, electric through the circuit

  1. When coil is used separately = 9.16A
  2. When coils are used in series = 4.58 A
  3. When coils are used in parallel = 18.33 A


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