In the figure ϵ1 = 3.05 V, ϵ2 = 0.996 V, R1 = 5.84 Ω, R2 = 2.44 Ω, R3 = 3.49 Ω, and both batte

Step 1: Calculate the current flowing through each resistor using Ohm's law, I = V/R.

In the figure ϵ1 = 3.05 V, ϵ2 = 0.996 V, R1 = 5.84 Ω, R2 =...

In the figure $\epsilon_1 = 3.05 \text{ V}$, $\epsilon_2 = 0.996 \text{ V}$, $R_1 = 5.84 \Omega$, $R_2 = 2.44 \Omega$, $R_3 = 3.49 \Omega$, and both batteries are ideal. What is the rate at which energy is dissipated in (a) $R_1$, (b) $R_2$, and (c) $R_3$? What is the power of (d) battery 1 and (e) battery 2?

Transcript

00:01 So, in this question we are given voltage of battery 1 which is being denoted by e1 equal to 2 .54 volts and voltage of battery 2 which is being denoted by e2 equal to 1 .04 volts and r1 equal to 5 .03 ohms and r2 equal to 1 .62 ohms and r3 equal to 4 .23 ohms.

01:11 Now according to kirchhoff's rn law i1 plus i2 equal to i3.

01:47 Now let's mark it as equation 1.

01:52 Also according voltage law we have the sum of the sum of the voltage in the closed circuit is zero.

02:54 Now consider the kirchhoff's napsat loop equation e1 minus r1 multiplied by i1 minus r3 multiplied by i3 equal to zero which implies now putting the values 2 .54 minus 5 .03 multiplied by i1 minus 4 .23 multiplied by i3 which is equal to zero.

04:15 Now let's mark it as equation.

04:22 Now consider the kirchhoff's right side loop equation e2 minus r2 multiplied by i2 minus r3 multiplied by i3 which is equal to zero.

05:06 Now putting the values in the equation 1 .04 minus 1 .62 multiplied by i2 minus 4 .23 multiplied by i3 which is equal to zero.

05:35 Now let's mark it as equation 3.

05:37 Now from equation 1 and 3 we have 1 .04 minus 1 .62 multiplied by i3 minus i1 minus 4 .23 multiplied by i3 which is equal to zero.

06:14 So after simplifying this equation we have 1 .04 plus 1 .62 multiplied by i1 minus 5 .85 multiplied by i3 which is equal to zero.

06:41 Now let's mark it as equation 4.

06:43 Now from equation 2 and 4 we have 2 .54 minus 5 .03 multiplied by i1 minus 4 .1 .04 plus 1 .62 multiplied by i1 divided by 5 .85 which is equal to zero.

07:41 So after simplifying this equation we have 1 .788 minus 6 .2 multiplied by i1 which is equal to zero.

08:02 So from here we get the value of i1 as 0 .2884 amperes...

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