a) uncover (^I_2). Express her answer usingtwo significant figures, in Amps.

You are watching: The ammeter in the figure reads 3.0 a. (figure 1)

b) find (^mathcalE). Express her answer usingtwo far-reaching figures, in Volts.

Concepts and reason

The concept used right here is Kirchhoff"s existing law and Kirchhoff"s voltage law. Firstly, apply Kirchhoff"s existing law come the left loop and also Kirchhoff"s voltage legislation to the best loop and use the equation derived from the loop to calculate the existing (I_2). In the 2nd part, calculate the worth of (varepsilon) by utilizing the worth of existing calculated in the previous part.

Fundamentals

Kirchhoff"s current law:

At any junction in circuit, the algebraic amount of currents will certainly be zero. Kirchhoff"s voltage law:

In any closed loop that circuit, the algebraic amount of voltages will be zero.

The following number shows the given circuit diagram.

In this circuit diagram, suggest A and B is shown, at point A both currents room coming and they will go v A come B.

a) The Kirchhoff"s existing law is offered at junction (mathrmA), it is created as, (I_1+I_2=I_A B ldots ldots) (1)

The current in between (mathrmA) and also (mathrmB), it way the present (I_A B) will certainly be same to the reading of fastened ammeter, which is

(3 mathrm~A). Therefore, that is composed as, (I_A B=3 mathrm~A)

The voltage difference in between (A) and (B) is composed as, (V_A B=I_A B R_A B)

Substitute (3 mathrm~A) for (I_A B) and also (2 Omega) because that (R_A B) in above expression.

\$\$ eginaligned V_A B &=(3 mathrm~A)(2 Omega) \ &=6 mathrm~V endaligned \$\$

Apply the Kirchhoff"s voltage legislation in loop 1 , it is written as, (9 mathrm~V-3 I_1=V_A B)

Substitute the value (6 mathrm~V) because that (V_A B) in over expression.

\$\$ eginarrayc 9 mathrm~V-3 I_1=6 mathrm~V \ 3 I_1=3 mathrm~V \ I_1=1 mathrm~A endarray \$\$

Substitute the worth (3 mathrm~A) because that (I_A B) and (1 mathrm~A) because that (I_1) in expression (1)

(1 mathrm~A+I_2=3 mathrm~A)

\$\$ I_2=2 mathrm~A \$\$

Part a The worth of (I_2) is (2 mathrm~A).

Apply the Kirchhoff"s current law to uncover the relation between currents, use the Kirchhoff"s voltage law and calculate (I_1) and also with the aid of (I_1), calculation the value (I_2).

(b) use the Kirchhoff"s voltage law in loop 2 , it is created as, (varepsilon-4.5 I_2=V_A B)

Substitute the value (6 mathrm~V) for (V_A B) and (2 mathrm~A) for (I_2) in above expression.

\$\$ eginarrayc varepsilon-4.5(2 mathrm~A)=6 mathrm~V \ varepsilon=15 mathrm~V endarray \$\$

Part b The worth of (varepsilon) is (15 mathrm~V).

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Apply the Kirchhoff"s voltage law and calculate (I_1) and also with the aid of (I_1), calculation the value (I_2). Use the Kirchhoff"s voltage legislation in loop 2 and calculate the (varepsilon).