1.

0.5 amp current is passing through the 3 bulbs with 4Omega,12Omegaand8Omega resistances and they are connected in series ? Find the voltage of the combination.

Answer»

Solution :I = 0.5 AMP `= 1/2` amp.
VOLTAGE at bulb - 1 `(V_(1))=IR_(1)=1//2xx4=2V.`
Voltage at bulb -2 `(V_(2))=IR_(2)=1//2xx12=6V.`
Voltage at bulb - 3 `(V_(3))=IR_(3)=1//2xx8=4V.`
Votage of the combination is `V=V_(1)+V_(2)+V_(3)`
= 2 V + 6 V + 4 V = 12 V.



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