Question

An ideal voltmeter V is connected to a 2.0 O resistor and a battery with
emf 5.0 V and internal resistance 0.5 O as shown in the figure on the
right. (a) What is the current in the 2.0 O resistor? (b) What is the terminal
voltage of the battery? (c) What is the reading on the voltmeter? Explain
your answers.
                    

Answer 1

Concepts and reason

The concept required to solve this problem is Ohm’s law and the terminal voltage of battery.

Initially, refer the circuit diagram given in the question. Later, calculate the current through the resistor and the terminal voltage of the battery. Finally, calculate the voltmeter reading.

Fundamentals

The expression for the Ohm’s law is as follows:

$V = IR$

Here, I is the current and R is the resistance.

The expression for the terminal voltage is as follows:

${V_{\rm{t}}} = \varepsilon - IR$

Here, $\varepsilon$ is the EMF.

The expression to calculate the voltmeter reading is as follows:

$V = {V_{\rm{t}}} - {V_{2.0{\rm{ }}\Omega }}$

Here, ${V_{\rm{t}}}$ is the terminal voltage and ${V_{2.0{\rm{ }}\Omega }}$ is the voltage of the resistor of resistance $2.0{\rm{ }}\Omega$

(a)

The equation $I = \frac{V}{R}$ indicates that the current is inversely proportional to the resistance, keeping the voltage constant. The voltmeter given the circuit diagram is an ideal voltmeter and the resistance of an ideal voltmeter is infinite.

Substitute $\infty$ for R in the equation $I = \frac{V}{R}$ .

$\begin{array}{c}\\I = \frac{V}{\infty }\\\\ = 0.00{\rm{ A}}\\\end{array}$

Therefore, the current through the $2.00{\rm{ }}\Omega$ resistor is equal to 0.00 A.

(b)

Substitute 5.00 V for $\varepsilon$ , 0.00 A for I, and $2.00{\rm{ }}\Omega$ for R in the equation ${V_{\rm{t}}} = \varepsilon - IR$ .

$\begin{array}{c}\\{V_{\rm{t}}} = 5.00{\rm{ V}} - \left( {0.00{\rm{ A}}} \right)\left( {2.00{\rm{ }}\Omega } \right)\\\\ = 5.00{\rm{ V}} - {\rm{0}}{\rm{.00 V}}\\\\{\rm{ = 5}}{\rm{.00 V}}\\\end{array}$

(c)

Substitute 5.00 V for ${V_{\rm{t}}}$ and IR for ${V_{{\rm{2}}{\rm{.00 }}\Omega }}$ in the equation $V = {V_{\rm{t}}} - {V_{2.0{\rm{ }}\Omega }}$ .

$V = 5.00{\rm{ V}} - \left( {IR} \right)$

Now, substitute 0.00 A for I and $2.00{\rm{ }}\Omega$ for R in the equation $V = 5.00{\rm{ V}} - \left( {IR} \right)$ .

$\begin{array}{c}\\V = 5.00{\rm{ V}} - \left( {0.00{\rm{ A}}} \right)\left( {2.00{\rm{ }}\Omega } \right)\\\\ = 5.00{\rm{ V}}\\\end{array}$

Ans: Part a

The current through the $2.00{\rm{ }}\Omega$ resistor is equal to 0.00 A.

Part b

The magnitude of the terminal voltage is equal to 5.00 V.

Part c

The reading of the voltmeter is equal to 5.00 V.

Answer 2

Current is zero

Answer 3

SOLUTION :

An ideal voltmeter has infinite resistance, so current in the given circuit = V / ∞ = 0 amp.

a.  Current through 2.0 Ω resistor = 0 amp. (ANSWER).

b.  Terminal voltage of the battery = 5.0 - 0 * 0.5 = 5.0 V (ANSWER).

c. Reading on the volt meter = terminal voltage of the battery - 0 * 2.0  = 5.0 V (ANSWER).