Consider the network of four resistors shown in the diagram, where R_1 = 2.00 rm Omega, R_2 = 5.00 rm Omega, R_3 = 1.00 rm Omega, and R_4 = 7.00 rm Omega. The resistors are connected to a constant voltage of magnitude V.

**Answer 1**

1) R1 and R2 create an equivalent resistance of

R12 = 1 /( 1/2 + 1/5)

R12 = 1.429 Ω

so the total resistance of the circuit is

R_A = R12 + R3 + R4

R_A = 1.429 + 1 + 7

R_A = 9.43 Ω

2) R1 and R6 make an equivalent resistance of

R16 = 2 + 3 = 5 Ω

R16 in parallel to R2 of 5 Ω gives an equivalent of

R162 = 1 / (1/5 + 1/5) = 2.5 Ω

when the switch is open R7 has no affect in the circuit so that R4 and R5 are in series with R162

R_B = R162 + R4 + R5

R_B = 2.5 + 7 + 3

R_B = 12.5 Ω

3) with the switch closed R4 and R7 become parallel and create an equivalent

R47 = 1 / (1/7 + 1/3) = 2.1

so that

R_C = R162 + R3 + R47 + R5

R_C = 2.5 + 1 + 2.1 + 3

R_C = 8.6 Ω

**Answer 2**

E = 50 V R1 = 3 R2 = 9 R3 = 3 R4 = 5 I1 = E / (R1 + R3) I1 = 50 / (3 + 3) I1 = 50 / 6 I1 = 8.33 A I2 = E / (R2 + R4) I2 = 50 / (9 + 5) I2 = 50 / 14 I2 = 3.fifty seven A modern however R1 and R3 = 8.33 Amps ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ modern by using R2 and R4 = 3.fifty seven Amps ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ It = 8.33 + 3.fifty seven It = 12 A equivalent Resistance, Re = E / It Re = 50 / 12 Re = 4.167 equivalent Resistance = 4.167 ? ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

**Answer 3**

Post the figure.

**Answer 4**

“the diagram” is where?

what does ” rm Omega” mean?

no wonder no one answered it.