Answer:
simple but some different you have too put value onlyyy
Explanation:
Given that E=8.0V,V=120V,r=0.5Ω,andR=15.5Ω
Current in the circuit during charging is given by
I=Total voltageTotal resistance=V−ER−r=120−815.5+05=11216=7Ω
During charging , the current flows inside the battery in a direction
opposite to that during discharge. Hence, the terminal voltage
of the battery during charging is
E+Ir=8+7×0.5=11.5V
A series resistor in the charging circuit limits the current drawn from the external source. in its absence, the current will be dangerously high.
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Answer:
simple but some different you have too put value onlyyy
Explanation:
Given that E=8.0V,V=120V,r=0.5Ω,andR=15.5Ω
Current in the circuit during charging is given by
I=Total voltageTotal resistance=V−ER−r=120−815.5+05=11216=7Ω
During charging , the current flows inside the battery in a direction
opposite to that during discharge. Hence, the terminal voltage
of the battery during charging is
E+Ir=8+7×0.5=11.5V
A series resistor in the charging circuit limits the current drawn from the external source. in its absence, the current will be dangerously high.