Engineering, 11.03.2020 03:21 leriscepowell9004

An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0205 Ω, and a 1.53-V carbon-zinc dry cell, whose internal resistance is 0.105 Ω. All four cells are connected in series as part of a single-loop circuit. The resistance of the rest of the circuit is 6.625 Ω.

(a) What current, 1, in amperes, flows through the toy's circuit?

(b) How much power, P, in watts, is supplied to the toy?

(c) The dry cell fails in such a way that it maintains its emf while its internal resistance increases greatly. That results in a reduced power of only 0.500 W supplied to the toy. What is the internal resistance, r2,failing, in ohms, of the failed dry cell?

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Answers

Answer from: od3008

(a) The current in amperes that flows through the toy's circuit is 0.923A

(b) The power supplied to the toy is 5.78721W

Explanation:

From the circuit diagram attached. We have the electric component:

B1 = 3* 1.58 = 4.74V

B2 = 1.53V

r1 = 3*0.0205 = 0.0615Ω

r2 = 0.105Ω

R = 6.625Ω

Since the internal resistances and the resistor R are connected in series, we can calculate the total resistance RT as

RT = r1 + r2 +R = 0.0615 + 0.105 + 6.625

= 6.7915Ω

Total Voltage supplied to the circuit by both batteries V = B1 + B2 = 4.74 + 1.53 = 6.27V

(a) CIRCUIT CURRENT

The current I, flowing through the circuit is i = $\frac{V}{R_{T}}$ = $\frac{6.27}{6.7915}$ =0.923A

The current in amperes that flows through the toy's circuit is 0.923A

(b) THE POWER SUPPLIED TO THE TOY

Power P = I*V =0.923*6.27 = 5.78721W

The power supplied to the toy is 5.78721W

Due to dry cell failure, the power supplied to the toy is reduced to 0.5W

Now Power P = $\frac{V^{2} }{R}$ . To calculate the new total resistance of the circuit we will make R the subject of the formular

R= $\frac{V^{2} }{P}$ = $\frac{6.27^{2} }{0.5}$ = $\frac{39.3129}{0.5}$ = 78.6258Ω

Remember that RT = r1 + r2 + R

r1 =RT- (R +r2)

r1 = 78.6258 - 6.6865 =71.9393Ω

The internal resistance r2 of the failed dry cell is 72Ω

Answer from: kita2911

Part a: The current flowing through the toy's circuit is 0.92 A.

Part b: The power of the toy is 5.60 W

Part c: The failing resistance is 16.42 Ω

Explanation:

From the given information

The 3 Alkaline Cells are given as E1=E2=E3=1.58 V

The internal resistance of the Alkaline Cells is R1=R2=R3=0.0205Ω

The Carbon Zinc dry cell is given as E4=1.53 V

The internal resistance of the Carbon Zinc dry cell is as R4=0.105Ω

Resistance of the Circuit is Rc=6.625 Ω

The circuit is given as in the indicated figure.

Part a

The current is given as

$I=\dfrac{V}{R}\\I=\dfrac{E_1+E_2+E_3+E_4}{R_1+R_2+R_3+R_4+R_c}$

By substituting the values in the equation

$I=\dfrac{E_1+E_2+E_3+E_4}{R_1+R_2+R_3+R_4+R_c}\\I=\dfrac{1.58+1.58+1.58+1.53}{0.0205+0.0205+0.0205+0.105+6.625}\\I=0.923 A$

The current flowing through the toy's circuit is 0.92 A.

Part b:

The power is given as

$P=I^2R_c\\P=0.92^2*6.625\\P=5.60 W$

So the power of the toy is 5.60 W

Part c:

Now the power is given as P=0.5W

The equivalent circuit is given in the attached picture.

So the value of I is calculated as

$P=I^2R_c\\I=\sqrt{\dfrac{P}{R_c}}\\I=\sqrt{\dfrac{0.5}{6.625}}\\I=0.27 A$

The equivalent voltage is given as

$P=\dfrac{V^2}{R_c}\\V=\sqrt{P\times R_c}\\V=\sqrt{0.5\times 6.625}\\V=1.82 V$

So now the resistance of the failure is given as

$R_i=\dfrac{E_1+E_2+E_3+E_4-V}{I}-[R_1+R_2+R_3]\\R_i=\dfrac{1.58+1.58+1.58+1.53-1.82}{0.27}-[0.0205+0.0205+0.0205]\\R_i=16.42 \Ohm$

So the failing resistance is 16.42 Ω

An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0

Answer from: Quest

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explanation:

Answer from: Quest

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An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0...