# The resistance R = 10.0 Ohms and ε1 = 13.3 V and ε2 = 18.7 V. What is the current through the Resistor R at the bottom of the circuit ?

### 4 Answers

- oldschoolLv 72 months ago
Let the left side be Node Vx and the right side = ground. Now write an equation for the currents leaving Vx = 0 by KCL

(Vx-13.3)/10 + (Vx-18.7)/10 + Vx/10 = 0

Multiply eq by 10

3Vx = 13.3+18.7 = 32

Vx = 32/3

(32/3-13.3)/10 = -79/300

(32/3-18.7)/10 = -241/300

32/3/10 = 320/300 = 16/15 A <<<<<

- KnrLv 42 months ago
for the circuit effective emf = 18.7 - 13.3 = 5.4 volt and effective resistance R’ = R/3 = 10/3 ohm

and so total current is I = 5.4 / (10/3) = 1.62 amp.

then, current through each resistor is i = 1.62 / 3 = 0.54 amp.

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- FiremanLv 72 months ago
Let the current from E2 is i2 & E1 is i1 & the current passing through the bottom resistance is i3, BY the kirchoff's Laws:

=>i1 + i2 = i3 ------------(i)

By the upper loop:

=>i1R - i2R = E1 - E2

=>13.3 - 18.7 = 10(i1 - i2)

=>i1 - i2 = -0.54 -----------------(ii)

By the lower loop:

=>18.7 = i2R + i3R

=>18.7 = 10(i2 + i1 + i2) {as i3 = i1 + i2 -----------(i)}

=>i1 + 2i2 = 1.87 ---------------(iii)

By 2 x (ii) + (iii):

=>3i1 = -2 x 0.54 + 1.87

=>i1 = 0.26 amp

By (ii:

=>i2 = 0.54 + 0.26

=>i2 = 0.80 amp

Thus by (i):

=>i3 = 0.26 + 0.80 = 1.06 amp

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- oubaasLv 72 months ago
E2 conducting

Re2 = R+R/2 = 1.5 R

I2' = 18.7/15 = 1.247 A

I1' = I3' = 1.247/2 A

E1 conducting

Re1 = R+R/2 = 1.5 R

I1'' = 13.3/15 = 0.887 A

I2'' = I3'' = 0.887/2 A

I3 = I3'+I3'' = 1.066 A

I1 = I1''-I1' = 0.887-1.247/2 = 0.263

I2 = I2'-I2'' = 1.247-0.887/2 = 0.803

V3 = I3*10 = 10.66 V

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Thank you!