| Direction: Q.21 to Q.25 |
| In physics, an electric power measure of the rate of electrical energy transfer by an electric circuit per unit time. It is denoted by P and measured using the SI unit of power is the watt or one joule per second. Electric power is commonly supplied by sources such as electric batteries and produced by electric generators. |
| The formula for electric power is given by P = VI |
| where, P is the power, V is the potential difference in the circuit, I is the electric current |
| Power can also be written as |
| \[P={{I}^{2}}R\] and \[P={{V}^{2}}/R\] |
| The above two expressions are got by using Ohm's law, where, voltage, current and resistance are related by the following relation |
| \[V=IR\] |
| where, R is the resistance in the circuit, V is the potential difference in the circuit, I is the electric current. |
| The given figure shows four bulbs 1, 2, 3 and 4, consume same power. The resistance of bulb 1 is \[36\,\Omega \]. |
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| Read the above passage carefully and give the answer of the following questions. |
A) \[4\text{ }\Omega \]
B) \[\text{9 }\Omega \]
C) \[\text{18 }\Omega \]
D) \[\text{12 }\Omega \]
Correct Answer: B
Solution :
(b) \[\text{9 }\Omega \] The bulbs 2 and 3 are in series, current through them is same. \[\therefore {{l}_{2}}={{l}_{3}}={{l}_{b}}\](say) Now, bulb 1 and combination of 2 and 3 are in parallel, \[\therefore {{V}_{1}}=({{V}_{2}}+{{V}_{3}})=V\](say) Since, all bulbs consumes same power. \[\therefore {{P}_{2}}={{P}_{3}}\Rightarrow l_{b}^{2}{{R}_{2}}=l_{b}^{2}{{R}_{3}}\] \[\Rightarrow {{R}_{2}}={{R}_{3}}\] So, \[{{V}_{2}}={{V}_{3}}=\frac{V}{2}\] Now, \[{{P}_{1}}=\frac{V_{1}^{2}}{{{R}_{1}}}=\frac{{{V}^{2}}}{36}\] And \[{{P}_{3}}=\frac{V_{3}^{2}}{{{R}_{3}}}=\frac{{{V}^{2}}}{4{{R}_{3}}}\] \[\because {{P}_{1}}={{P}_{3}}\] \[\therefore \frac{{{V}^{2}}}{36}=\frac{{{V}^{2}}}{4{{R}_{3}}}\Rightarrow {{R}_{3}}=9\,\Omega \]
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