Question

A circuit consisting of three ideal batteries with voltages E1, E2, and E3, and three ideal resistors with resistances R1, R2, and R3, is shown in the figure. Calculate the current Ip through point P. Let the sign of the current correspond to its direction, with "up" being positive. + Ej = 32.5 V, E2 = 21.0 V, E 3 = 34.5 V R1 = 5.20 k2, R2 = 19.5 k-2, R3 = 4.50 k12 co + Ip = MA

Answer 1

total voltage =$E_1+E_2+E_3=32.5+21.0+34.5=88 volts$

total resistance =$R_1+R_2+R_3=(5.2+19.5+4.5)k\Omega=29.2k\Omega$

current = $\frac{voltage}{resistance }=\frac{88}{29.2\times10^3}=3.01\times10^{-3}ampere =3.01mA$