(a) How would you reduce the circuit to an equivalent single resistor connected

Question

(a) How would you reduce the circuit to an equivalent single resistor connected to the battery? Use this procedure to find the equivalent resistance of the circuit.


(b) Find the current delivered by the battery to this equivalent resistance.


(c) Determine the power delivered by the battery.


(d) Determine the power delivered to the 50.0- resistor.

Four resistors are connected to a battery with a terminal voltage of 12 V, as shown in the figure below. (Assume R 31.0 n and R 73.0 n.) 50.0 20.0 (2 12 V

Explanation / Answer

a) first R1 and 50 ohm in series so we can add them,

R1 = 31 ohm , let R3 =50 ohm

so total is 81 ohm.

now 80 ohm and R2 73 ohm are parallel,

so 1/R = 1/80+1/73 = > R= 38.18 ohm

This 38.18 ohm is now series with 20 ohm so totlly 58.18 ohm.

so equalent resistance is 58.18 ohm.

b) Current I =V / Requalent =>12/58.18= 0.2062 A

c) Power P =V x I => 12 x 0.2062 =>2.4744 W

d) voltage drop at 20 ohm is V= 0.2062 x 20 = 4.124

Voltage in parallel connection= 12V-4.124V=7.876V

7.876= I (31+50)
I = 97.234mA
7.876= 97.234mA (31+50)
=3.014V+4.861V
where: voltage in 50ohm resistor= 4.861V
use formula P= VI=4.861(97.234mA)=472.6544 m watts

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