The circuit shown in the figure below contains two real batt

The circuit shown in the figure below contains two \"real batteries\", each having significant One of the batteries has an EMF of 6.00 V and an internal resistance of 2.00 ohms. The other battery has an EMF of 3.00 V and an internal resistance of 1.00 ohm. The circuit also contains a 3.00 ohm resistor \"R\", as shown in the Figure. The current in the 3.00 ohm resistor equals 1.50 A. 2.00 A. 3.00 A. 0.500 A. 1.00 A. The potential difference across the terminals of the 6 V battery from point \"a\" to point \"b equals 6.00 V. 7.00 V. 5.00 V. 4.00 V. 2.00 V. The potential difference across the terminals of the 3 V battery from point \"c\" to point \"b\" equals 3.00 V. 3.50 V. 2.50 V. 4.00 V. 2.00 V.

Solution

let current be I

So 6-3 V = (r1+r2+R)I

I = 3/6 =.5 A

so current 21 is c.

22)

6-I*r 1= 5 V

23) 3+I.r2 = 3.5 V