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you are given two circuits with two batteries of emf e and internal resistance r1 each. circuit a has the batteries connected in series with a resistor of resistance r2, and circuit b has the batteries connected in parallel to an equivalent resistor.

Respuesta :

Answer:

In which direction does the current in circuit A flow?

counterclockwise

What is the power dissipated by the resistor of resistance R2 for circuit A, given that E=10 V, R1=300ohms, and R2=5000ohms?

Calculate the power to two significant figures.

0.064W

For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?

R1/R2 = 1

Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B?

Some answer choices overlap; choose the most restrictive answer.

R2>R1

Explanation:

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Ver imagen ACYee
onyebuchinnaji

The current flowing through the circuit A is [tex]I = \frac{e}{r_1 + r_2}[/tex].

"Your question is not complete, it seems to be missing the following information",

find the the current flowing through the resistors in circuit A;

The given parameters;

  • emf of each battery = e
  • internal resistance of circuit A = r₁
  • resistance of circuit = r₂

The equivalent resistance of the circuit A in series is calculated as follows;

[tex]R_e = r_1 + r_2[/tex]

The emf of the battery connected to circuit A

emf = e

The current flowing through the circuit A is calculated as follows;

[tex]V= IR\\\\I = \frac{V}{R} \\\\I = \frac{e}{R_e} \\\\I = \frac{e}{r_1 + r_2}[/tex]

Thus, the current flowing through the circuit A is [tex]I = \frac{e}{r_1 + r_2}[/tex].

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