
Two resistors 400 Ω and 800 Ω are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 k Ω across 400 Ω resistors is close to:
A. 2.05 V
B. 2 V
C. 1.95 V
D. 1.8 V
Answer
161.4k+ views
Hint:In the series combination of resistance, the current is same for each resistor. A resistor is known to be connected in series if same amount of current flows through the resistors. We can determine the potential difference of the circuit by using ohm’s law.
Formula used
By ohm’s law,
$V=IR$
Where V is voltage, I is current and R is resistance.
Resistors in series,
\[\begin{array}{l}{R_s} = {R_1} + {R_2}....\\{\rm{ }}\end{array}\]
Resistors in series,
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}....\]
Complete step by step solution:
To calculate the effective resistance of the circuit as:

Image:Circuit diagram
As 400 Ω and 10 k Ω are connected in parallel, thus resistance along AB will be,
Let \[{R_1} = 400\Omega \] and \[{R_2} = 10k\Omega = 1000\Omega \] are two resistance connected in parallel as shown in figure, then by using formula of parallel combination of resistance,
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\\ \Rightarrow \dfrac{1}{{{R_p}}} {\rm{ = }}\dfrac{1}{{400}} + \dfrac{1}{{1000}}\\ \Rightarrow \dfrac{1}{{{R_p}}} {\rm{ = }}\dfrac{{400 \times 1000}}{{400 + 1000}}\\ \Rightarrow {\rm{ }}{{\rm{R}}_p}{\rm{ = }}\dfrac{{5000}}{{13}}\Omega \]
Now this resistance is connected with the given resistance \[{R_3} = \] 800 Ω as shown in figure, then by using formula of series combination of resistance,
\[{R_s} = {R_p} + {R_3}\\ \Rightarrow {R_s}{\rm{ = }}\dfrac{{5000}}{{13}} + 800\\ \Rightarrow {R_s}{\rm{ = }}\dfrac{{15400}}{{13}}\Omega \]
To find current after connecting a battery of 6 V,
As, \[I = \dfrac{V}{R} = \dfrac{6}{{\dfrac{{15400}}{{13}}}}\]
\[I= \dfrac{{39}}{{7700}}A\]
As we know
\[V = IR\\ \Rightarrow V{\rm{ = }}\dfrac{{39}}{{7700}} \times \dfrac{{5000}}{{13}}\\ \therefore V{\rm{ = 1}}{\rm{.95 V}}\]
Therefore the potential difference measured by a voltmeter of 10 k Ω across 400 Ω resistors is close to 1.95 V.
Hence, option C is the correct answer
Note:A resistor is defined as a passive two terminal electrical component that implements electrical resistance as a circuit element. It reduces the flow of current and lower voltage levels within the circuits.
Formula used
By ohm’s law,
$V=IR$
Where V is voltage, I is current and R is resistance.
Resistors in series,
\[\begin{array}{l}{R_s} = {R_1} + {R_2}....\\{\rm{ }}\end{array}\]
Resistors in series,
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}....\]
Complete step by step solution:
To calculate the effective resistance of the circuit as:

Image:Circuit diagram
As 400 Ω and 10 k Ω are connected in parallel, thus resistance along AB will be,
Let \[{R_1} = 400\Omega \] and \[{R_2} = 10k\Omega = 1000\Omega \] are two resistance connected in parallel as shown in figure, then by using formula of parallel combination of resistance,
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\\ \Rightarrow \dfrac{1}{{{R_p}}} {\rm{ = }}\dfrac{1}{{400}} + \dfrac{1}{{1000}}\\ \Rightarrow \dfrac{1}{{{R_p}}} {\rm{ = }}\dfrac{{400 \times 1000}}{{400 + 1000}}\\ \Rightarrow {\rm{ }}{{\rm{R}}_p}{\rm{ = }}\dfrac{{5000}}{{13}}\Omega \]
Now this resistance is connected with the given resistance \[{R_3} = \] 800 Ω as shown in figure, then by using formula of series combination of resistance,
\[{R_s} = {R_p} + {R_3}\\ \Rightarrow {R_s}{\rm{ = }}\dfrac{{5000}}{{13}} + 800\\ \Rightarrow {R_s}{\rm{ = }}\dfrac{{15400}}{{13}}\Omega \]
To find current after connecting a battery of 6 V,
As, \[I = \dfrac{V}{R} = \dfrac{6}{{\dfrac{{15400}}{{13}}}}\]
\[I= \dfrac{{39}}{{7700}}A\]
As we know
\[V = IR\\ \Rightarrow V{\rm{ = }}\dfrac{{39}}{{7700}} \times \dfrac{{5000}}{{13}}\\ \therefore V{\rm{ = 1}}{\rm{.95 V}}\]
Therefore the potential difference measured by a voltmeter of 10 k Ω across 400 Ω resistors is close to 1.95 V.
Hence, option C is the correct answer
Note:A resistor is defined as a passive two terminal electrical component that implements electrical resistance as a circuit element. It reduces the flow of current and lower voltage levels within the circuits.
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