Chapter 23, Problem 54A

a.

To determine

The reading of ammeter connected in the given circuit.

Answer to Problem 54A

The ammeter reads the current flowing through the circuit and it shows a value $0.2\text{ A}$.

Explanation of Solution

Given:

A circuit consists of two resistors of value, $35\Omega ,15\Omega$ connected in series to a 10-V battery. The voltage and current in the circuit are measured using meters as shown in the circuit.

Formula Used:

The ammeter in a circuit measures the current through the element in series with it. Here, the ammeter measures the total current through the circuit which can be calculated using Ohm’s law as,

$\begin{array}{l}V=IR\\ I=\frac{V}{R}\end{array}$

Calculation:

Consider the potential divider circuit shown in Figure 1.

Figure 1

Here, the supply voltage is 10-V and the two resistors have values $35\Omega$ and $15\Omega$ . The ammeter is connected in series to the supply voltage and it measures the total current in the circuit. It can be obtained as,

$\begin{array}{c}I=\frac{10}{35+15}\\ =0.2\text{A}\end{array}$

Conclusion:

The value read by the ammeter is 0.2 A.

b

To determine

The reading of voltmeter 1 connected across a given resistance.

Answer to Problem 54A

The voltmeter 1 reads a voltage of value $\text{7 V}$.

Explanation of Solution

Given:

A circuit consists of two resistors of value, $35\Omega ,15\Omega$ connected in series to a 10-V battery. The voltage and current in the circuit are measured using meters as shown in the circuit.

Formula Used:

The voltmeter connected across a resistor reads the voltage drop across it. Here, for a given supply voltage and resistors in series, the voltage across each of them can be calculated using the voltage division rule as,

$V_{R2}=V\frac{R_2}{R_1+R_2}$

Calculation:

Consider the potential divider circuit shown in Figure 1. Here, the supply voltage is 10-V and the two resistors have values, $35\Omega ,15\Omega$ . The voltmeter 1 is connected across $35\Omega$ . It measures the voltage across the resistor and is given by,

$\begin{array}{c}{V}_{35\Omega }=10\frac{35}{35+15}\\ =7\text{ V}\end{array}$

Conclusion:

The reading of voltmeter 1 is 7 V.

c.

To determine

The reading of voltmeter 2 connected across a given resistance.

Answer to Problem 54A

The voltmeter 2 reads a voltage of value $\text{3 V}$.

Explanation of Solution

Given Information:

A circuit consists of two resistors of value, $35\Omega ,15\Omega$ connected in series to a 10-V battery. The voltage and current in the circuit are measured using meters as shown in the circuit.

Formula Used:

The voltmeter connected across a resistor reads the voltage drop across it. Here, for a given supply voltage and resistors in series, the voltage across each of them can be calculated using the voltage division rule as,

$V_{R2}=V\frac{R_2}{R_1+R_2}$

Calculation:

Consider the potential divider circuit shown in Figure 1. Here, the supply voltage is 10-V and the two resistors have values, $35\Omega ,15\Omega$ . The voltmeter 2 is connected across $15\Omega$ . It measures the voltage across the resistor and is given by,

$\begin{array}{c}{V}_{15\Omega }=10\frac{15}{35+15}\\ =3\text{ V}\end{array}$

Conclusion:

The reading of voltmeter 2 is 3 V.

d.

To determine

The energy supplied by the battery in a given time.

Answer to Problem 54A

The energy supplied by the battery in one minute is $120\text{ J}$.

Explanation of Solution

Given:

A circuit consists of two resistors of value, $35\Omega ,15\Omega$ connected in series to a 10-V battery. The voltage and current in the circuit are measured using meters as shown in the circuit.

Formula used:

The energy supplied by a source E can be expressed in terms of the power dissipation and the time for which power is drawn, t, as

  $E=Pt$

The power drawn by the circuit is a function of its resistance R and current flowing through it, I which can be expressed as,

  $P=I^{2}R$

Calculation:

Since the circuit has two resistors in series, the equivalent resistance value is

  $\begin{array}{c}{R}_{eq}=35+15\\ =50\Omega \end{array}$

Substituting the current and resistance value, the power can be obtained as,

  $\begin{array}{c}P={0.2}^2\times 50\\ =2\text{ W}\end{array}$

The energy supplied is to be calculated for a duration of one minute. Now, substituting the value of power and time, the energy can be calculated as,

  $\begin{array}{c}E=2\times 1\times 60\\ =120\text{ W/s}\\ \text{=120 J}\end{array}$

Conclusion:

The energy supplied by the battery for a given time is 120 J.

e.

To determine

The equivalent resistance of the circuit.

Answer to Problem 54A

The equivalent resistance of the circuit is $\text{50 }\Omega$.

Explanation of Solution

Given:

A circuit consists of two resistors of value, $35\Omega ,15\Omega$ connected in series to a 10-V battery. The voltage and current in the circuit are measured using meters as shown in the circuit.

Formula used:

The equivalent resistance of two resistors in a series circuit is given by,

  $R_{eq}=R_1+R_2$

Calculation:

Substituting the individual resistance values, the equivalent resistance value is

  $\begin{array}{c}{R}_{eq}=35+15\\ =50\Omega \end{array}$

Conclusion:

The equivalent resistance of the circuit is $50\Omega$ .