91影视

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Resistor one =\(224 - \Omega \)

Resistor two = \(589 - \Omega \)

Both the resistor connected by = \({\rm{90}}{\rm{.0 - V}}\)

Reading by resistor \(224 - \Omega \)= \({\rm{23}}{\rm{.8 V}}\)

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The passive two terminal electrical component by which electrical resistance get implemented as a circuit element is known as resistor R.

\(\begin{aligned}{}{R_{Total}} = {R_1} + {R_2} + {R_3} + ...(series)\\{R_{Total}} &= \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + ...(parallel)\\R &= \frac{V}{I}\\V &= voltage\\I &= current\end{aligned}\)

The current of the resistor will be,

\(\begin{aligned}{}I &= {I_1} &= {I_2}\\I &= \frac{{90.0{\rm{V}}}}{{{R_1} + {R_2}}}\\I &= \frac{{90.0{\rm{V}}}}{{224\Omega + 589\Omega }}\\I &= 0.1107{\rm{A}}\end{aligned}\)

So the voltage in the resistor will be,

\(\begin{aligned}{}{V_1} &= {I_1}{R_1}\\{V_1} &= \left( {0.1107{\rm{A}}} \right)\left( {224\Omega } \right)\\{V_1} &= 24.8{\rm{V}}\end{aligned}\)

And,

\(\begin{aligned}{}{V_2} &= {I_1}{R_1}\\{V_2} &= \left( {0.1107{\rm{A}}} \right)\left( {589\Omega } \right)\\{V_2} &= 65.2{\rm{V}}\end{aligned}\)

So the voltage across 1^st and 2^nd resistor will be respectively \({V_1} = 24.8{\rm{V}}\)and \({V_2} = 65.2{\rm{V}}\). \(\)

The following figure is describing the resistor network,

As given in the question the potential difference across the voltmeter terminals is \({\rm{23}}{\rm{.8 V}}\), so to find out the current flow in the system we have to apply the following ohm鈥檚 law,

The voltage drop across the \(589 - \Omega \) resistor is \(90.0{\rm{V}} - 23.8{\rm{V = 66}}{\rm{.2V}}\), so now,

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\(\begin{aligned}I &= \frac{V}{R}\\I &= \frac{{{\rm{66}}{\rm{.2V}}}}{{589\Omega }}\\I &= 0.1124{\rm{A}}\end{aligned}\)

The voltage drop across resistor \(224 - \Omega \)= \({\rm{23}}{\rm{.8 V}}\), so now,

\(\begin{aligned}{I_2} = \frac{{{\rm{23}}{\rm{.8V}}}}{{224\Omega }}\\{I_2} = 0.1062{\rm{A}}\\{I_1} = I - {I_2}\\{I_1} = 0.1124{\rm{A}} - 0.1062{\rm{A = }}0.0062{\rm{A}}\end{aligned}\)

The resistor will be,

\(\begin{aligned}{R_V} = \frac{V}{{{I_1}}}\\{R_V} = \frac{{{\rm{23}}{\rm{.8V}}}}{{0.0062{\rm{A}}}}\\{R_V} = 3840\Omega \end{aligned}\) \(\)

The voltmeter resistance is \({R_V} = 3840\Omega \) \(\)

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The circuit that is connected with the voltmeter is shows as below,

\(\begin{aligned}{l}\\\end{aligned}\)

\(\begin{aligned}\frac{1}{{{R_{eq}}}} = \frac{1}{{3840\Omega }} + \frac{1}{{589\Omega }}\\{R_{eq}} = 510.7\Omega \end{aligned}\)

The voltmeter current,

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\(\begin{aligned}I &= \frac{V}{R}\\I &= \frac{{{\rm{90}}{\rm{.0V}}}}{{224\Omega + 510.7\Omega }}\\I &= 0.1225{\rm{A}}\end{aligned}\)

The potential drop across the first resistor is

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\(\begin{aligned}{V_1} &= {I_1}{R_1}\\{V_1} &= \left( {0.1225{\rm{A}}} \right)\left( {224\Omega } \right)\\{V_1} &= 27.4{\rm{V}}\end{aligned}\)

So on the second one it will be,

\(\begin{aligned}{}\\\end{aligned}\)

\(90{\rm{V}} - 27.4{\rm{V = 62}}{\rm{.6V}}\)

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No, any real voltmeter will draw some current and thereby reduce the current through the resistance whose voltage is being measured. Thus the presence of the voltmeter connected in parallel with the resistance lowers the voltage drop across that resistance. The resistance of the voltmeter in this problem is only about a factor of ten larger than the resistances in the circuit, so the voltmeter has a noticeable effect on the circuit.