The ideal battery in Figure a has emf 61 V Plot 1 in Figure

The ideal battery in Figure (a) has emf = 6.1 V. Plot 1 in Figure (b) gives the electric potential difference V that can appear across resistor 1 of the circuit versus the current j in that resistor. The scale of the V axis is set by V_s = 19.3 V, and the scale of the j axis is set by j_s = 3.13 mA. Plots 2 and 3 are similar plots for resistors 2 and 3, respectively. What is the current in resistor 2?

Solution

R = V/i

from the V,I given

R3 = 6.43V/3.13mA = 2.06 kOhms

R2 = 12.87/3.13e-3 = 4.11 kOhms

R1 = 19.3/3.13e-3 = 6.17 kOhms

R1 and R2 are in parallel result of R1 and R2

           = 6.17*4.11/(6.17+4.11) = 2.47 kOhms

voltage across R1 and R2 = 6.1*2.47/(2.47+2.06) = 3.32 V

current R2 = 3.32/4.11e+3 = 0.81 mA