1. A(n) 700- and a(n) 1800- resistor are connected in series with a 12-V battery

Question

1. A(n) 700- and a(n) 1800- resistor are connected in series with a 12-V battery. (my professor didnt give any figure, sorry :(

Part A

What is the voltage across the 1800- resistor?

2.

Part A

Determine the equivalent resistance of the "ladder" of equal 175- resistors shown in (Figure 1) . In other words, what resistance would an ohmmeter read if connected between points A and B?

Express your answer to three significant figures and include the appropriate units.

art B

What is the current through resistor a if a 50.0-V battery is connected between points A and B?

Part C

What is the current through resistor b?

Express your answer to three significant figures and include the appropriate units.

Part D

What is the current through resistor c?

Express your answer to three significant figures and include the appropriate units

Explanation / Answer

Given,

R1 = 700 Ohm ; R2 = 1800 Ohm ; V = 12 V

A)Since the resistirs are connected in series, the equivalent resistance would be:

Req = R1 + R2

Req = 700 + 1800 = 2500 Ohm

The current in the circuit is:

I = V/Req = 12/2500 = 0.0048 A

Its a fact that the same current flows through the resistors connected in series. So the drop across 1800 will be:

V2 = I2 R2

V2 = 0.0048 x 1800 = 8.64 Volts

Hence, V2 = 8.64 Volts

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