Question

Is the unknown element a battery or a resistor?

A. The unknown element is a battery.
B. The unknown element is a resistor.
C. Whether it is a battery or a resistor cannot be determined.

  potential differences must be zero. Part C Although Kirchhoffs junction law is needed only when there are one or more junctions in a circuit, Kirchhoffs loop law is used for analyzing any type of circuit, as explained in the following tactics box. What is the absolute value of the potential difference AV across the unknown element in the circuit shown in (Figure 3)? Express your answer in volts. AV1.0 V Figure 3 of 3 Submit Hints My Answers Give Up Review Part 3.0 v Correct Part D Is the unknown element a battery or a resistor? 6.0 V The unknown element is a battery. The unknown element is a resistor. Whether it is a battery or a resistor cannot be determined 4.0 V Submit My Answers Give Up

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Answer #1
Concepts and reason

The concept used to solve this problem is potential difference in the circuit.

Initially, the absolute value of the unknown potential difference can be calculated by using the net potential in the circuit. Later, the unknown element is whether a battery or a resistance can be determined by calculating the value of the unknown element

Fundamentals

“The change in the potential developed in the circuit is state as voltage. If the battery connected to a circuit, the positive charged particles flows across the resistors and capacitors in the circuit the Kirchhoff’s loop rule is developed”.

“According to the Kirchhoff’s loop states the conservation of energy that the sum of all voltages in a closed loop has zero”.

The expression for the net potential in the circuit is,

V1+V2+VxV3=0{V_1} + {V_2} + {V_x} - {V_3} = 0

Here, V1,V2,V3{V_1}\;,{V_2}\;,{V_3}are the potential connected in the circuit andVx{V_x}is the unknown potential.

(c)

The expression for the unknown potential is,

V1+V2+VxV3=0{V_1} + {V_2} + {V_x} - {V_3} = 0

Substitute 4V4\;{\rm{V}}forV1{V_1}, 3V3\;{\rm{V}}forV2{V_2}, 6V6\;{\rm{V}}forV3{V_3}

(4V)+(3V)+Vx(6V)=0Vx1VVx=1V\begin{array}{l}\\\left( {4\;{\rm{V}}} \right) + \left( {3{\rm{V}}} \right) + {V_x} - \left( {6{\rm{V}}} \right) = 0\\\\{V_x} - 1\;{\rm{V}}\\\\|{V_x}| = 1\;{\rm{V}}\\\end{array}

(d)

The expression for the unknown element is,

V1+V2+VxV3=0{V_1} + {V_2} + {V_x} - {V_3} = 0

Substitute 4V4\;{\rm{V}}forV1{V_1}, 3V3\;{\rm{V}}forV2{V_2}, 6V6\;{\rm{V}}forV3{V_3}

(4V)+(3V)+Vx(6V)=0Vx1VVx=1V\begin{array}{l}\\\left( {4\;{\rm{V}}} \right) + \left( {3{\rm{V}}} \right) + {V_x} - \left( {6{\rm{V}}} \right) = 0\\\\{V_x} - 1\;{\rm{V}}\\\\|{V_x}| = 1\;{\rm{V}}\\\end{array}

The calculated value is the potential of the battery. Thus the connected element in the circuit is a battery.

Ans: Part c

The absolute value of the unknown potential is1V1\;{\rm{V}}.

Part d

The unknown element is battery.

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