Q79P A circuit consists of two batter... [FREE SOLUTION]

Chapter 19: Q79P (page 803)

A circuit consists of two batteries (with negligible resistance), six ohmic resistors and connecting wires that have negligible resistance. The resistance $R_{1}$ is $10 惟$ , $R_{2}$ is $20 惟$ , $R_{3}$ is $30 惟$ , $R_{4}$ is $12 惟$ , $R_{5}$ is $15 惟$ and $R_{6}$ is $20 惟$ . Unknown currents $I_{1}$ , $I_{2}$ , $I_{3}$ , $I_{4}$ , $I_{5}$ and $I_{6}$ have their directions marked on the circuit diagram in figure 19.87.
(a) Write down a set of equations that could be solved for the six unknown currents. Make sure you can explain how to you got these equations. (b) When a correct set of equations is solved the currents are as follows (to the nearest miiampeares) $I_{1} = 0.4394 A$ , $I_{2} = 0.3312 A$ , $I_{3} = 0.0065 A$ , $I_{4} = 0.1082 A$ , $I_{5} = 0.3247 A$ and $I_{6} = 0.4329 A$ . Check your equations by substituting in these numbers. (c) Suppose that you connect the negative lead of a voltmeter to location C. What does the voltmeter read, including both magnitude and sign? (d) What does the power output of the 5 V battery? (e) Resistor is made of a very thin metal wire that is 3 mm long, with a diameter of 0.1 mm. What is the electric field inside the metal resistor.

Short Answer

Expert verified

The set of equations for unknown currents are $I_{1} = I_{2} + I_{4}, I_{2} = I_{3} + I_{5}$ , $I_{1} = I_{3} + I_{6}$ , $20 - 10 I_{1} - 15 I_{4} - 12 I_{6} - 20 I_{1} = 0$ , $5 + 15 I_{4} - 20 I_{2} = 0$ and $- 5 - 30 I_{3} + 12 I_{6} = 0$ .

Step by step solution

Identification of given data

The potential of the battery for loop 1 is $V_{1} = 20 V$

The potential of the battery for loop 2 and loop 3 is $V_{2} = 5 V$

The resistance of the first resistor is $R_{1} = 10 惟$

The resistance of the second resistor is $R_{2} = 20 惟$

The resistance of the third resistor is $R_{3} = 30 惟$

The resistance of the fourth resistor is $R_{4} = 12 惟$

The resistance of the fifth resistor is $R_{5} = 15 惟$

The resistance of the sixth resistor is $R_{6} = 20 惟$

Conceptual Explanation

The conservation of charge at every node and conservation of potentials in every loop is used to solve the above problem. The conservation of charge says that the incoming currents at a node are equal to the outgoing current from that node. The conservation of potential says that the net potential in any loop is always zero.

Determination of equations to find the unknown currents

Apply the Kirchoff鈥檚 current law at C.

$I_{1} = I_{2} + I_{4}$

Apply the Kirchoff鈥檚 current law at D.

$I_{2} = I_{3} + I_{5}$

Apply the Kirchoff鈥檚 current law at F.

$I_{1} = I_{3} + I_{6}$

Apply the Kirchoff鈥檚 voltage law in the loop 1.

$\begin{array}{rcl} V_{1} - I_{1} R_{1} - I_{4} R_{5} - I_{6} R_{4} - I_{1} R_{6} & = & 0 \\ 20 - 10 I_{1} - 15 I_{4} - 12 I_{6} - 20 I_{1} & = & 0 \end{array}$

Apply the Kirchoff鈥檚 voltage law in the loop 2.

$\begin{array}{rcl} V_{2} + I_{4} R_{5} - I_{2} R_{2} & = & 0 \\ 5 + 15 I_{4} - 20 I_{2} & = & 0 \end{array}$

Apply the Kirchoff鈥檚 voltage law in the loop 3.

$\begin{array}{rcl} - V_{2} - I_{3} R_{3} + I_{6} R_{4} & = & 0 \\ - 5 - 30 I_{3} + 12 I_{6} & = & 0 \end{array}$

Therefore, the set of equations for unknown currents are $I_{1} = I_{2} + I_{4}$ , $I_{2} = I_{3} + I_{5}$ , $I_{1} = I_{3} + I_{6}$ , $20 - 10 I_{1} - 15 I_{4} - 12 I_{6} - 20 I_{1} = 0$ , $5 + 15 I_{4} - 20 I_{2} = 0$ and $- 5 - 30 I_{3} + 12 I_{6} = 0$ .