a) uncover \$$^I_2\$$. Express your answer usingtwo far-reaching figures, in Amps.

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

b) uncover \$$^\\mathcalE\$$. Express her answer usingtwo significant figures, in Volts.

Concepts and reason

The concept used right here is Kirchhoff\"s existing law and also Kirchhoff\"s voltage law. Firstly, apply Kirchhoff\"s present law come the left loop and Kirchhoff\"s voltage legislation to the best loop and use the equation acquired from the loop to calculation the existing \$$I_2\$$. In the second part, calculate the value of \$$\\varepsilon\$$ by making use of the worth of present calculated in the ahead part.

Fundamentals

Kirchhoff\"s current law:

At any kind of 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 certainly be zero.

The following number shows the offered circuit diagram.

In this circuit diagram, allude A and B is shown, at allude A both currents space coming and also they will certainly go with A come B.

a) The Kirchhoff\"s existing law is provided at junction \$$\\mathrmA\$$, the is written 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 be same to the analysis of enclosed ammeter, which is

\$$3 \\mathrm~A\$$. Therefore, the is created as, \$$I_A B=3 \\mathrm~A\$$

The voltage difference in between \$$A\$$ and \$$B\$$ is written as, \$$V_A B=I_A B R_A B\$$

Substitute \$$3 \\mathrm~A\$$ because that \$$I_A B\$$ and \$$2 \\Omega\$$ for \$$R_A B\$$ in above expression.

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

Apply the Kirchhoff\"s voltage law in loop 1 , that is written as, \$$9 \\mathrm~V-3 I_1=V_A B\$$

Substitute the value \$$6 \\mathrm~V\$$ because that \$$V_A B\$$ in above expression.

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

Substitute the worth \$$3 \\mathrm~A\$$ because that \$$I_A B\$$ and \$$1 \\mathrm~A\$$ for \$$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 existing law to find the relation in between currents, use the Kirchhoff\"s voltage law and also calculate \$$I_1\$$ and also with the help of \$$I_1\$$, calculation the worth \$$I_2\$$.

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

Substitute the worth \$$6 \\mathrm~V\$$ because that \$$V_A B\$$ and \$$2 \\mathrm~A\$$ for \$$I_2\$$ in above expression.

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

Part b The worth of \$$\\varepsilon\$$ is \$$15 \\mathrm~V\$$.

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Apply the Kirchhoff\"s voltage law and also calculate \$$I_1\$$ and also with the help of \$$I_1\$$, calculation the value \$$I_2\$$. Apply the Kirchhoff\"s voltage law in loop 2 and calculate the \$$\\varepsilon\$$.