Consider the circuit shown in the figure below. (Assume R 1 = 11.0 , R 2 = 1.95

Question

Consider the circuit shown in the figure below. (Assume R1 = 11.0 , R2 = 1.95 , and V = 7.30 V.)

(a) Calculate the equivalent resistance of the R1 and 5.00- resistors connected in parallel.


(b) Using the result of part (a), calculate the combined resistance of the R1, 5.00- and 4.00- resistors.


(c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00- resistor.


(d) Combine the equivalent resistance found in part (c) with the R2 resistor.


(e) Calculate the total current in the circuit.
A

(f) What is the voltage drop across the R2 resistor?
V

(g) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00- resistor.
V

(h) Calculate the current in the 3.00- resistor.
A

Explanation / Answer

Solution:

Part a) Equivalent Resistance= (1/R1 + 1/5) ^-1 = (1/11 + 1/5) ^-1 =3.44 ohms ( Ans)

Part b) Equivalent Resistance = 3.44 ohms+ 4 ohms= 7.44 ohms (Ans) As they are in series now.

Part c) Equivalent Resistance = (1/7.44+ 1/3)^-1 = 2.14 ohms (Ans)

Part d) Equivalent Resistance= 2.14 ohms+ 1.95 ohms = 4.09 ohms (Ans)

Part e) Total Current (I) = V/ R = 7.3/4.09= 1.78 Amperes (Ans)

Part f) V= IR = 1.78* 1.95 =3.48 Volts (Ans)

Part g) Voltage across 3ohm resistor= 7.3 V- 3.48 V= 3.82 Volts ( Ans)

Part h) I= V/R = 3.82/3= 1.27 Amperes ( Ans)

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Good luck!:)

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