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Answered by
11
Given:
Internal resistance = r=0.5 ohms
Resistance connected in series= R1= 15.5 ohms
D.c supply = 120 v
Total effective resistance = R = r + R1
R= 0.5 + 15.5
= 16 ohms
Current = I= v/R
I = 120/ 16= 7.5 AMPERES.
CURRENTdue to 8 volts battery = 8/ 16= 0.5A
Therefore, net current = 7.5 - 0.5 =7A
Therefore, voltage across internal resistance =7× 0.5 = 3.5 V
Voltage across terminal of battery is = 8v + 3.5 volts
= 11.5 volts
Internal resistance = r=0.5 ohms
Resistance connected in series= R1= 15.5 ohms
D.c supply = 120 v
Total effective resistance = R = r + R1
R= 0.5 + 15.5
= 16 ohms
Current = I= v/R
I = 120/ 16= 7.5 AMPERES.
CURRENTdue to 8 volts battery = 8/ 16= 0.5A
Therefore, net current = 7.5 - 0.5 =7A
Therefore, voltage across internal resistance =7× 0.5 = 3.5 V
Voltage across terminal of battery is = 8v + 3.5 volts
= 11.5 volts
thank u very much mam
its absolutely correct
Answered by
112
Explanation:
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