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In the given circuit, the voltage difference across the\[20K\Omega \] resistor is
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A) $3.76V$
B) \[27.6V\]
C) $36V$
D) \[37.6V\]

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Last updated date: 13th Jun 2024
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Answer
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Hint:To solve the given problem check the diagram given in the question. They have asked to find the voltage difference across the \[20K\Omega \] resistor. Consider the position of that particular resistor and remember the formula that relates the voltage, current and resistor value.

Formula used:
Ohm’s law,
\[{\text{V = IR}}\], here $V$= potential difference, $I$= current flows in the circuit, $R$= resistance across the resistors
Series combination of resistors: $R = {R_1} + {R_2} + .............. + {R_n}$
Parallel combination of resistors: $\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + .............. + \dfrac{1}{{{R_n}}}$

Complete step by step answer:
Given details:
\[{R_1} = 20K\Omega \] ,
\[{R_2} = 5K\Omega \],
\[{R_3} = 8K\Omega \],
\[{R_4} = 6K\Omega \],
\[\;{R_5} = 5K\Omega \],
\[V = 9V\]
From the circuit diagram, ${R_1}$ and ${R_2}$ are connected in series combination,
Therefore,
${R'} = {R_1} + {R_2}$
After substituting the resistor values in the given equation use addition to add the values as it is a series combination,
$ \Rightarrow (20 + 5)$
$ \Rightarrow 25\Omega $ and
From the circuit diagram, ${R_3}$ and ${R_4}$ are connected in series combination,
Therefore,
${R^{''}} = {R_3} + {R_4}$
After substituting the resistor values in the given equation use addition to add the values as it is a series combination,
$ \Rightarrow ({\text{8 + 6)}}$
$ \Rightarrow 14$
Again $R'$ and ${R^{''}}$ are connected in parallel combination,
Therefore, we can calculate as,
$\dfrac{1}{{{{\text{R}}^{'''}}}} = \dfrac{1}{{{{\text{R}}'}}} + \dfrac{1}{{{{\text{R}}^{''}}}}$
We can use fraction addition to add the values, we get,
$ \Rightarrow \dfrac{1}{{{{\text{R}}^{'''}}}} = \dfrac{1}{{25}} + \dfrac{1}{{14}}$
$ \Rightarrow \dfrac{1}{{{{\text{R}}^{'''}}}} = 0.04 + 0.071428$
With the help of addition again we get,
$ \Rightarrow $$\dfrac{1}{{{{\text{R}}^{'''}}}} = 0.111428$
We need ${R^{''}}$only we are taking the left-hand side numerator to the right hand side and dividing it with the $0.111428$ we get,
$ \Rightarrow {R^{'''}} = 8.97440\Omega $
Now, the current flows through the right-hand side circuit having $9V$ voltage drop can be calculated as,
${{\text{I}}'} = \dfrac{9}{{8.97440}}$
By using division, we get,
$ \Rightarrow 1.002852A$
Now, the current flows through the left-hand side circuit having $9V$ voltage drop can be calculated as,
${I^{''}} = \dfrac{9}{5}$
By using division, we get,
$ \Rightarrow 1.8A$
According to Ohm’s law $V = IR$. We have got the values for $I$and $R$ we can substitute in the formula to find the voltage.
We can calculate the voltage across \[20K\Omega \]
 $ \Rightarrow V = 1.8 \times 20$
$\therefore 36V$

The correct option is C

Note:
-Fixed resistors: It has resistances that change slightly with time, temperature or operating voltage.
-Variable resistors: It can be used to adjust circuit elements (such as a volume control or a lamp dimmer), or as sensing devices for heat, humidity, light, force, or chemical activity.
-Series resistance: It is the combination of resistors connected in series.
-Parallel resistance: It is the combination of resistors connected in parallel.
-Ohm’s law: This law states that potential difference across a circuit is directly proportional to the current flows through the circuit.