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Consider the circuit shown below V=18.00V. The terminal voltage of the battery is.
(a) Find the equivalent resistance of the circuit.
(b) Find the current through each resistor.
(c) Find the potential drop across each resistor.
(d) Find the power dissipated by each resistor.
(e) Find the power supplied by the battery.

Consider The Circuit Shown Below V1800V The Terminal Voltage Of The Battery Is A Find The Equivalent Resistance Of The Circuit B Find The Current Through Each R class=

Sagot :

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a) R= R1 + R2 + R3
So,
4 + 1 + 4= 9
b) I= V/R
So,
So,
18/9= 2 A
c) V= I x R
d) P= I*2 x R

I don’t know how to do the last two.

(a) The equivalent resistance of the circuit is 9 Ω.

(b)  The current through each resistor is 2A.

(c)  The potential drop across each resistor is V₁ =8V = V₃ and V₂ =2 V.

(d) The power dissipated by each resistor is  P₁ =P₃ =16W and  P₂ = 4 W.

(e)  The power supplied by the battery is 36W.

What is equivalent resistance?

The resistances when connected in series has the equivalent resistance equal to the sum of all the individual resistances.

(a) In the given circuit, the equivalent resistance is

Req= R₁ + R₂ + R₃

Req = 4 + 1 + 4

Req= 9 Ω

The equivalent resistance is 9 Ω.

b) The current through each resistor will be same

I= V/R = 18/9= 2 A

c) The potential drop across each resistor.

V₁= IxR₁ = 2x 4 =8Volts = V₃

V₂ = IR₂ =2x1 =2 Volts

(d) The power dissipated by each resistor

P₁ =P₃ = V₁²/R₁

= 8² / 4 =16 Watt

and P₂ = V₂²/R₂

P₂ = 2²/1 = 4 Watt

(e) The power supplied by the battery.

P = I² x Req

P = 2² x 9

P = 36 Watt

Learn more about equivalent resistance.

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