Electrical devices in parallel

Electricity and Magnetism Level 4

The supply voltage to a room is $\SI {120}{\volt}$ . The resistance of the lead wires is $\SI {6}{\ohm}$ . A $\SI {60} {\watt}$ bulb is already switched on. What is the decrease of voltage across the bulb, when a $\SI {240}{\watt}$ heater is switched on in parallel to the bulb? if ur answer is V, give [V], [.] represents the greatest integer function.

Note: The combined resistance of the lead wires is $\SI{6}{\ohm}$ (since there are two of them).

Here supply voltage is taken as rated voltage.

The answer is 10.

2 solutions

A Former Brilliant Member
Mar 14, 2017

Resistance of bulb = 120 120 240 60

× = Ω Resistance of Heater = 120 120 60 240

× = ΩVoltage across bulb before heater is switched on, 1 120 V 240 246

= × Voltage across bulb after heater is switched on, 2

120 V 48

54 = ×

Decrease in the voltage is V1 − V2 = 10.04 (approximately) Note: Here supply voltage is taken as rated voltage.

Ye kon bolega ke supply voltage is taken as rated voltage? cbse wale hai he pagal

Prakhar Bindal - 4 years, 2 months ago

muje pagal kah ra k bhai ?

A Former Brilliant Member - 4 years, 2 months ago

bhai physics(cbse2017) k last Q ka answer infinity h kya ?

A Former Brilliant Member - 4 years, 2 months ago

@A Former Brilliant Member – Set kon sa tha?

Prakhar Bindal - 4 years, 2 months ago

@Prakhar Bindal – sabse muskil wwala -3

A Former Brilliant Member - 4 years, 2 months ago

@A Former Brilliant Member – mera 2 tha medium wala . sabse asan 1 tha

Prakhar Bindal - 4 years, 2 months ago

@Prakhar Bindal – haan 2 thoda assan tha 3 sabse muskil ( meri hi aisi thaisi karni thi :P) us q ka to btade wo sabka same tha , how much r u expecting ?

A Former Brilliant Member - 4 years, 2 months ago

@A Former Brilliant Member – Nhi bhai last kon sa tha? bread lene gya tha :P

Prakhar Bindal - 4 years, 2 months ago

Abe ye mains ka question unko pagal kah rha hu

Prakhar Bindal - 4 years, 2 months ago

Resha Dwika Hefni Al-Fahsi
Aug 22, 2017

First Event: When the heater has not been connected then the circuit on the lamp will be like the picture on the side so that:

V = I. R 120 = I. (R + 6) 120 I = I^2. R + 6 I^2 120 I = 60 + 6 I^2 I^2- 20 I+ 10 = 0

So the possibility of I = 10 + 3√10 A or I = 10 - 3√10 A

If, I = 10 + 3√10 A then:

P = V I 60 = V x 10 + 3√10 V = 3.07 V, so R = 0.158 Ω

If, I = 10 - 3√10 A then: P = V I 60 = V x 10 - 3√10 V = 116.9 V, so R= 227.841 Ω

Assumed If the lamp has a high voltage and resistance it is obtained I = 10 - 3√10 A

Second Event:

The second occurrence when the heater is connected in parallel with the lamp then the circuit will be like the picture on the side so that:

V1= V2 √ (P1 / R1) = √ (P2 / R2) √ (60 / 227,841) = √ (240 / R2) R2 = 56.96 Ω

So that the combined resistance of Rtotal = (56.96 x 227.841) / (56.96 + 227.841) = 45.568 Ω Then the series current will change by:

V = I R 120 = I 51.568 I = 2.327 A

Thus, the current on the lamp will also change by:

I1 = R2/ (R1+ R2) × I = 56.96 / (227.841 + 56.96) × 2.327 = 0.465 A

Thus, the new lamp voltage is: V1 = I1 R1 = 0.465 × 227.841 = 106.9 V

Then the voltage drop in the lamp will be approximately 116.9 - 106.9 = $10 V$

Electrical devices in parallel

The answer is 10.

2 solutions

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