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A battery of emf 2v and internal resistance of 10 ohm is connected in series to 10 ohm resistors and a parallel combination of resistors of resistance 20 ohm and X. The battery derives a current 0.5 A to the circuit. What is the resistance X and current Y in the circuit.
Ans; X=3ohm, Y=0.33A how?
aimanmoin123:
I want to attach a image its not working. 3 resistors 10 , 6 and X are in parallel
Where 10 ohm resistor is itself the combination of a 9ohm and 1ohm resistor in series.
Okay
Answers
Answered by
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If the No load EMF of Battery is 2 Volts, at a current of 0.5 Amps, the terminal voltage of the Battery will fall by 10x0.5 = 5 Volts i.e. the terminal voltage from battery at 0.5 Amps current will be 1.5 Volts.
In this case it is simply R=V/I ; Where V, the Battery Voltage
and I, the current in the circuit = 0.5 Amps.
So the resistance = 1.5/0.5 ohms = 3 ohms. ( x = 3 ohms )
I know this may not help u but I tried my best to answer....
sorry for incomplete answer
Answered by
1
Hey mate......
here's Ur answer.....
Ques))) A battery of emf 2v and internal resistance of 10 ohm is connected in series to 10 ohm resistors and a parallel combination of resistors of resistance 20 ohm and X. The battery derives a current 0.5 A to the circuit. What is the resistance X and current Y in the circuit.
Ans; X=3ohm, Y=0.33A how?
Solution =
R = V/1
I = 0.5 A
Resistañce = 1.5 / 0.5
= 3 ohm
for 2nd part I think so the given answer value is wrong..
Hope it helps♥
here's Ur answer.....
Ques))) A battery of emf 2v and internal resistance of 10 ohm is connected in series to 10 ohm resistors and a parallel combination of resistors of resistance 20 ohm and X. The battery derives a current 0.5 A to the circuit. What is the resistance X and current Y in the circuit.
Ans; X=3ohm, Y=0.33A how?
Solution =
R = V/1
I = 0.5 A
Resistañce = 1.5 / 0.5
= 3 ohm
for 2nd part I think so the given answer value is wrong..
Hope it helps♥
How did you get 1.5 as voltage?
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