A projector lamp can be used at a maximum voltage of 60 V, its resistance is 20 Ω , the series resistance (in ohms) required to operate the lamp from a 75 V supply is

Question

A projector lamp can be used at a maximum voltage of 60 V, its resistance is 20 Ω , the series resistance (in ohms) required to operate the lamp from a 75 V supply is (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

Current flowing through the lamp, I=(V/R)=(60/20)=3A The current in the lamp must remain 3A, when it is operated at 75 V supply. Let R be the resistance connected in series, then (75/20+R)=3 ⇒ 75=60+3R or 3R=75-60 or 3R=15 or R=5Ω

Q. A projector lamp can be used at a maximum voltage of 60 V, its resistance is 20 $Ω$ , the series resistance (in ohms) required to operate the lamp from a 75 V supply is

2

3

4

5

Current flowing through the lamp, $I = \frac{V}{R} = \frac{60}{20} = 3 A$ The current in the lamp must remain 3A, when it is operated at 75 V supply. Let R be the resistance connected in series, then $\frac{75}{20 + R} = 3$ $\Rightarrow$ $75 = 60 + 3 R$ or $3 R = 75 - 60$ or $3 R = 15$ or $R = 5Ω$

Q. A projector lamp can be used at a maximum voltage of 60 V, its resistance is 20 Ω , the series resistance (in ohms) required to operate the lamp from a 75 V supply is

2

3

4

5

Current flowing through the lamp, I=RV​=2060​=3A The current in the lamp must remain 3A, when it is operated at 75 V supply. Let R be the resistance connected in series, then 20+R75​=3 ⇒ 75=60+3R or 3R=75−60 or 3R=15 or R=5Ω

Q. A projector lamp can be used at a maximum voltage of 60 V, its resistance is 20 $Ω$ , the series resistance (in ohms) required to operate the lamp from a 75 V supply is

Current flowing through the lamp, $I = \frac{V}{R} = \frac{60}{20} = 3 A$ The current in the lamp must remain 3A, when it is operated at 75 V supply. Let R be the resistance connected in series, then $\frac{75}{20 + R} = 3$ $\Rightarrow$ $75 = 60 + 3 R$ or $3 R = 75 - 60$ or $3 R = 15$ or $R = 5Ω$