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When switch S in the figure is open, the voltmeter V of the battery reads 3.08...

When switch S in the figure is open, the voltmeter V of the battery reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.98 V, and the ammeter A reads1.64 A. Assume that the two meters are ideal, so they don't affect the circuit.

uploaded imagea) Find the emf.

b) Find the internal resistance of the battery

c) Find the circuit resistance R.

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Answer #1
Concepts and reason

The concept related to solve this problem is the internal resistance and terminal voltage. The concept of internal resistance is useful in analyzing many types of the electrical circuits and this concept applies to all kinds of electrical sources. First calculate the Emf of the battery. After that, calculate the internal resistance of the battery. Finally, calculate the resistance in the circuit.

Fundamentals

The resistance within a battery or other voltage source is internal resistance. Internal resistance causes a drop in the source voltage when there is a current. Electromotive force is the voltage developed by any source electrical energy.

The net resistance of the circuit when the resistor R and the internal resistance r are connected in series is as follows:

Rnet=R+r{R_{{\rm{net}}}} = R + r

Here, R is the resistance of the resistor and r is the internal resistance.

According to the ohm’s law, the potential difference is as follows:

V=IRV = IR

Here, V is the potential and I is the current.

The expression for the terminal voltage is as follows:

Vt=εIr{V_{\rm{t}}} = \varepsilon - Ir

Here, ε\varepsilon is the emf.

The current in the circuit is as follows:

I=εR+rI = \frac{\varepsilon }{{R + r}}

(a)

The current passing through the circuit when switch is in open position is equal to zero.

I=0I = 0

Substitute 00forIIin the equationV=εIrV = \varepsilon - Irand solve for ε\varepsilon .

V=ε(0)rε=V\begin{array}{c}\\V = \varepsilon - \left( 0 \right)r\\\\\varepsilon = V\\\end{array}

Substitute 3.08 V forVV.

ε=V=3.08V\begin{array}{c}\\\varepsilon = V\\\\ = 3.08\;{\rm{V}}\\\end{array}

The open circuit voltmeter reading is equivalent to the emf of the battery. Hence, emf of the battery is equal to 3.08 V.

(b)

The expression for the terminal voltage is as follows:

Vt=εIr{V_{\rm{t}}} = \varepsilon - Ir

Rewrite the equation to solve for r.

r=εVtIr = \frac{{\varepsilon - {V_{\rm{t}}}}}{I}

Substitute 3.08 V for ε\varepsilon , 2.98 V for Vt{V_{\rm{t}}}, 1.64 A for I.

r=3.08V2.98V1.64A=0.1V1.64A=0.061Ω\begin{array}{c}\\r = \frac{{3.08{\rm{ V}} - 2.98{\rm{ V}}}}{{1.64{\rm{ A}}}}\\\\ = \frac{{0.1{\rm{ V}}}}{{1.64{\rm{ A}}}}\\\\ = 0.061{\rm{ }}\Omega \\\end{array}

(c)

The current in the circuit is equal to,

I=εR+rI = \frac{\varepsilon }{{R + r}}

Rewrite this equation and solve for the circuit resistance.

R=εIrR = \frac{\varepsilon }{I} - r

Substitute 3.08 V forε\varepsilon , 1.64 A for I, and0.061Ω0.061{\rm{ }}\Omega for r.

R=3.08V1.64A0.061Ω=1.817Ω\begin{array}{c}\\R = \frac{{3.08{\rm{ V}}}}{{1.64{\rm{ A}}}} - 0.061{\rm{ }}\Omega \\\\ = 1.817{\rm{ }}\Omega \\\end{array}

Ans: Part a

The emf of the battery is 3.08 V.

Part b

The internal resistance of the battery is equal to 0.061Ω0.061{\rm{ }}\Omega .

Part c

The circuit resistance is equal to 1.817Ω1.817{\rm{ }}\Omega .

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