Question

An ideal battery creates a potential difference at the electrodes equal to its emf no matter what is attached. However, in reality, the electrochemical cells that creates the emf has its own internal resistance. In the circuit below the dashed box represents a real battery, and Ry is the internal resistance. It causes the battery output voltage (that you would measure between the top and bottom of the box) to be different from the emf. In this circuit, the emf is 9 V, the internal resistance is 212, R2 = 152, R2 = 1022, and R3 = 2012 SW1 RB R2 R1 E w RE What is the battery's output voltage in V with the switch open? Question 2 1 pts What is the battery's output voltage in V with the switch closed?

could you please help me with this question? Thank you :)

Answer 1

I Rai I = 0.529 A 17 sal T Question of Here switch is open so no corrent Jolow through R2 and Rz. So we eliminated R2 and Rizo Now we get the circuit Tatal seesistance of the aircuit R, R= Roth, = (2+15) R=172 9 Now Caerent blow in this circuit I= R Now battery's output voltage V= &- IRp → V= 9– 0.529% 2 = 7.94 V Battery cutput voltage with the sowitch open 7.94 V Question 2 Now the scuitch is closed. Here Rą and Rz are in Series R2 Connection. So equivalent resistance R₃ R4 = Rat R3 → R = 10+ 20 = 30.2 Now RA and R, are in parallel. So equivalent resistance Ř= Rp kap >> +žo + R= 102 E Now corrent blow in this circait I= RBTR =I= 0.75 A New battery's output voltage VE E-IRB 5 V= 9-0.75x2 = 7.5 Battery output voltage with the switch closed 7.5V RB Ri I 9 2+10