Question

Problem 19.57 The batteries shown ii the circuit in the figure (Figure 1) have negligibly small internal resistances. Assume that E = 85. 0 V and R = 17.6 OhmPart B Find the current through the 17.6 Ohm resistor. Part C Find the current through the 8.50 V battery.

Answer 1

Here ,

B)

let the current in 17.6 ohm is I ,

Now, current in 30 ohm is I1 ,

Using ohm's law

E = I1 * R

8.50 = I1 * 30

I1 = 0.283 A ,

Now, using ohm's law .

5 - 17.6 * I - 8.50 = 0

I = -3.5/17.6

I = -0.2 A

the current in the R is 0.2 A

C)

Now, current in battery is

Ib = I1 + I

Ib = 0.2 + 0.283

Ib = 0.483 A

the current in the battery is 0.483 A