Refer to the figure v1 110 omega r1 i 120 v r2 440 omega i2...

Refer to the figure. $V_1 = ?$ $1.10 \Omega$ $R_1$ $I = ?$ $12.0 V$ $R_2$ $4.40 \Omega$ $I_2 = ?$ $I_3$ $R_3$ $12.0 \Omega$ $V_p = V - V_1 = ?$ (a) Calculate $P_3$ (in W). Enter a number $W$ (b) Find the total power (in W) supplied by the source. $W$

Transcript

00:01 So for this question we have these three identical resistors.

00:06 So we've got r1, then r2, and then r3.

00:12 And they're all connected in series.

00:13 So this is r1, this is r2, and this is r3.

00:16 And we learn that the voltage across r1 is 10 volts, and the voltage across r3 is 36 volts.

00:24 And we also learn that the total power is 70 watts, and the current is 1 .4 amps.

00:31 So for the first part, we need to work out the value of the vt, and therefore the voltage drop across r2.

00:40 Well, we know that power is equal to the current multiplied by the voltage.

00:47 So therefore, our 70 watts is equal to 1 .4 amps multiplied by the total voltage.

00:54 So the total voltage is 70 divided by 1 .4.

00:58 So 70 divided by 1 .4 is 50.

01:03 So there are 50 volts in total.

01:06 So therefore, the voltage drop across r2 is going to equal 50 volts minus the voltage drop across r1, which is 10 volts, and the voltage drop across r3, which is 36 volts.

01:28 So that's 50 minus 46, which is 4 volts.

01:33 So it's 4 volts across r2.

01:35 Then for part b, we have to work out the equivalent resistance across the voltage source and the values of r1, r2, and r3.

01:53 So across our voltage source, well, our resistance r is equal to v.

02:00 So we know that v is equal to i times r.

02:03 The voltage is the current times by the resistance.

02:05 So therefore, r is equal to v over i.

02:10 And our voltage is across the entire thing is 50 volts.

02:16 And the current is 1 .4.

02:18 So 50 divided by 1 .4 is 35 .7 ohms.

02:31 And then our values for the resistance of r1, r2, and r3 will be as follows.

02:36 Well, we know for r1, it's going to be v1 over i.