Solving Electricity Prob w/ R1=7\Omega, R2=5\Omega, R3=4\Omega

[faoltaem]

The values for the resistors are: R1=7$$\Omega$$, R2=5$$\Omega$$, R3=4$$\Omega$$ (see diagram for placement of the resistors)
1) Suppose there is a current of 0.6A going through R1 and that the voltage supplied by the battery is 9V, determine the value of R4
2) Using the information above, determine the current through R3 and the voltage across R4

1) $$\frac{1}{R_{eq(3+4)}}$$ = $$\frac{1}{4}$$ + $$\frac{1}{R4}$$

= $$\frac{R4}{4R4}$$ + $$\frac{4}{4R4}$$

= $$\frac{R4 + 4}{4R4}$$
therefore R$$_{eq(3+4)}$$ = $$\frac{4R4}{R4 + 4}$$

R$$_{tot}$$ = R1 + R2 + R$$_{eq(3+4)}$$
15 = 7 + 5 + $$\frac{4R4}{R4 + 4}$$
$$\frac{4R4}{R4 + 4}$$ = 3
4R4 = 3(R4 + 4)
= 3R4 + 12
therefore R4=12$$\Omega$$

2) I = $$\frac{\epsilon}{R}$$ = 9/15 = 0.6A
V = IR = 0.6 x 12 = 7.2V

Is this correct?

 

[andrevdh]

1 is fine.

2. is a bit of a problem.

note that the statement says that the current through R1 is 0.6 A. This means that 0.6 amps flows through R2 and also through the parallel combination. So less than 0.6 amps should flow through R3.

Calculate the potential drop over the parallel combination. This same voltage appears over R3 and R4. Use this voltage and the known resistance of R3 to calc the current through it.

 

[faoltaem]

part 2 retry - electricity

P = VI = $$\frac{V^{2}}{R}$$ = I$$^{2}$$
I$$_{1+2}$$ = 0.6A
R = 7+5 = 12$$\Omega$$
P = 0.6$$^{2}$$ x 12 = 4.32W

V$$^{2}$$ = PR = 4.32 x 12 = 51.84V
V = 7.2V

P = I$$^{2}$$ x 16 = 4.32
I$$^{2}$$ = $$\frac{4.32}{16}$$ = 0.2A
I = 0.5A

I = $$\frac{\epsilon}{R}$$

$$\epsilon$$ = IR = 0.5 x 12 = 6V

 

[andrevdh]

?

It seems you like to do things the hard (and also hard to comprehend) way ... You should somehow word what you are doing when you are solving problems and label calculated quantities, like $$I_3$$, in order for your teacher to evaluate your problems in the future.

A simpler/my approach would be:

The total resistance of the parallel combination is three ohm as you have discovered yourself. This means that the voltage drop over the combination (and also the individual resistors of the parallel connection - R3 and R4) will be

$$V_P = 0.6 \times 3\ volts$$

This will be the potential difference over R4 (and R3), so it is quite simple to calc the current through R3 now with the resistance value of R3 given an the potential drop over it.