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| 1. |
Two bulbs are rated (P_1, V)"and"(P_2, V) . If they are connected (i) in series and (ii) in parallel across a supply V, find the power dissipated in the two combinations in terms of P_1 and P_2. |
| Answer» <html><body><p></p>Solution :Resistance of given bulbs are `R_1 = (V^2)/(P_1) " and " R_2 = (V^2)/(P_2)` <br/> (i) In series grouping net resistance `R_s = R_1 + R_2`. If `P_s` be the power dissipated in series arrangement then `R_s = (V^2)/(P_s)` <br/>`rArr (V^2)/(P_s) = (V^2)/(P_1) + (V^2)/(P_2) rArr (1)/(P_s) = (1)/(P_1) + (1)/(P_2) rArr P_s = (P_1 P_2)/(P_1 + P_2)`<br/>(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) In <a href="https://interviewquestions.tuteehub.com/tag/parallel-1146369" style="font-weight:bold;" target="_blank" title="Click to know more about PARALLEL">PARALLEL</a> grouping <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> resistance `R_P = (V^2)/(P_P)` is given as <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> relation :<br/>`(1)/(R_P) = (1)/(R_1) + (1)/(R_2) rArr (P_P)/(V^2) = (P_1)/(V^2) + (P_2)/(V^2) rArr P_P = P_1 + P_2`</body></html> | |