Four resistors are connected to a battery as shown in the figure. The current in the…
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Question “Four resistors are connected to a battery as shown in the figure. The current in the…”
Four resistors are connected to a battery as shown in the figure. The current in the battery is I, the battery emfis E= 6.90 V, and the resistor values are R1-R, R2 = 2R, R3 = 4R, R4-3R. Find the voltages across each resistor. Number R,-2R Number Number Number
Answer
Answer 1
The net resistance of the circuit is -R R2+R3R4 V 6.90 Volt.2.3V voltage drop across VIR 2.3V voltage drop across RS4-6.90-2.34.6V.4.6Volt.2.3 Volt.2.3 Volt.2.3 Volt.2.3 Volt.3.3 Volt.2.3 Volt.3.3 Volt.2.3 Volt.2.3 Volt.3.3 Volt.2.3 Volt.2.3 Volt. Voltage drop across R2-V2-IR2-x2R1.5333V.
Answer 2
SOLUTION :
Equivalent of all resistors connected to the battery
= R1 + (1 / (1/(R2 + R3) + 1/R4)
= R + ( 1/(2R + 4R) + 1/(3R) )
= R + 1/( ( 1 + 2) / (6R))
= R + 1/(1/(2R))
= R + 2R
= 3R
So, current flow from the battery, I
= V / (3R) = 6.90 / (3R)
= 2.30/ R amps.
Voltage , v1, across resistor R1
= I * R
= 2.30 / R * R
= 2.30 volts .
So, voltage across resistors (R2 + R3)
= 6.90 – 2.30 = 4.60 volts.
But, R2 + R3 = 2R + 4R = 6R
So, current through (R2 + R3)
= 4.60 / (6R)
= 0.7667 / R
So, voltage v2 across R2
= 0.7667 / R * 2R
= 1. 533 volts
So, voltage across R3
= 4.60 – 1.533
= 3.067 volts (ANSWER).
Voltage across R4
= Voltage across (R2 + R3)
= 4.60 volts (ANSWER).
So,
v1 = 2.30 volts
v2 = 1.533 volts
v3 = 4.067 volts
v4 = 4.60 volts
(ANSWER).
Conclusion
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