Suppose the circuit shown below has an internal resistance r = 0.730 Ohms. Inter

Question

Suppose the circuit shown below has an internal resistance r = 0.730 Ohms. Internal resistance manifests only when current is flowing through the battery. You can treat it as an additional resistor r in series with the battery. However, the voltage drop across the internal resistance occurs inside the battery, so the potential difference between the battery terminals decreases when current is flowing.

(a) What is the potential difference across the battery when the switch is open?

(b) Refer to the previous problem. What is the potential difference across the battery terminals after the switch has been closed for a long time?

I'm very confused about these knids of questions, hope someone can help me to slove!!!!!!!!!

Explanation / Answer

a) when the switch is open current is not flowing so potential drop across the battery is 9v

b)

when switch is closed

after long time capacitor acts as a open circuit

current=9/(11+5.6+0.73)=0.5193 A

so potential difference across battery=9-0.5193*0.73=8.62V

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