In the figure shown, battery `1` has emf`=6V` and internal resistance `= 1Omega.` Battery `2` has emf`=2V` and internal resistance `=3Omega.` The wires have negligible resistance. What is the potential difference across the teminals of battery `2`?

A shortage battey with an emf of 10V and an internal resistance 1Omega is connected across an extenal resistance R and liberates of 9W power. Find the potential difference across the teminals of the battery and the value or values of R.

A battery having e.m.f. 5 V and internal resistance 0.5 Omega is connected with a resistance of 4.5 Omega then the voltage at the terminals of battery is

A battery of emf 2V and internal resistance 0.5(Omega) is connected across a resistance in 1 second?

In Fig. 27-51. battery 1 has emf epsi_(1)= 12.0 V and internal resistance r_1= 0.025 Omega and battery 2 has emf_(2)= 12.0 V and internal resistance r_2= 0.012Omega . The batteries are connected in series with an external resistance R. (a) What R value makes the terminal-to-terminal potential difference of one of the batteries zero? (b) Which battery is that?

A battery having emf 4 V and internal resistance 0.5 Omega is connected with a resistance of 4.5 Omega . The voltage at the terminals of the battery now is

Two batteries one of emf 18 V and internal resistance 2Omega . And the other of emf 12 V and internal resistance 1Omega , are connected as shown. The voltmeter V will velocity record a reading of:

Two batteries one of the emf3V, internal resistance In and the other of emf l SV, internal resistance 2Omega are connected in series with a resistanceR as shown. If the potential difference between points a and b is zero, the resistance R is :

Two batteries one of the emf 3V , internal resistance 1Omega and the other of emf 15 V , internal resistance 2Omega are connected in series with a resistance R as shown. If the potential difference between points a and b is zero, the resistance R in Omega is