The equivalent resistance of the group of resistances is R. If another resistance is connected in parallel to the group, its new equivalent resistance becomes ${R_1}$ and if it is connected in series to the group, its new equivalent resistance becomes ${R_2}$. We have:
A. ${R_1} > R$
B. ${R_1} < R$
C. ${R_2} > R$
D. ${R_2} < R$
Answer
641.7k+ views
Hint: A circuit is known to be connected in series when the same amount of current flows through each of the resistors. While a circuit is said to be connected in parallel when voltage is same across each of the resistors. To find the solution of the given question assume another resistance ‘r’ added to the group and apply the formula of resistors in series and parallel.
Formula Used:
Resistors connected in series,
${R_{total}} = {R_1} + {R_2} + {R_3}$
Resistors connected in parallel,
$\dfrac{1}{{{R_{total}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$
Complete answer:
It is given that the equivalent resistance of the group of resistances is ‘R’. Let us assume that another resistance ‘r’ is added to this group.
Case 1: When ‘r’ is connected in parallel to the group
So, equivalent resistance ‘${R_1}$’ is given as,
${R_1} = \dfrac{{Rr}}{{R + r}}$
Thus, the equivalent resistance ‘${R_1}$’ is less than the equivalent resistance of the group of resistances ‘R’. i.e. ${R_1} < R$.
Case 2: When ‘r’ is connected in series to the group
So, equivalent resistance ‘${R_2}$’ is given as,
${R_2} = R + r$
Thus, the equivalent resistance ‘${R_2}$’ is greater than the equivalent resistance of the group of resistances ‘R’. i.e. ${R_2} > R$.
Therefore, it can be concluded that,
${R_1} < R$
${R_2} > R$
Hence, both option (B) and (C) are correct answers.
Note:
A resistor is defined as the passive two-terminal electrical component which implements electrical resistance as a circuit element. Resistors are used to reduce the current flow and lower voltage levels within the circuits. Most circuits often contain more than one resistor in order to limit the flow of charges in a circuit.
Formula Used:
Resistors connected in series,
${R_{total}} = {R_1} + {R_2} + {R_3}$
Resistors connected in parallel,
$\dfrac{1}{{{R_{total}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$
Complete answer:
It is given that the equivalent resistance of the group of resistances is ‘R’. Let us assume that another resistance ‘r’ is added to this group.
Case 1: When ‘r’ is connected in parallel to the group
So, equivalent resistance ‘${R_1}$’ is given as,
${R_1} = \dfrac{{Rr}}{{R + r}}$
Thus, the equivalent resistance ‘${R_1}$’ is less than the equivalent resistance of the group of resistances ‘R’. i.e. ${R_1} < R$.
Case 2: When ‘r’ is connected in series to the group
So, equivalent resistance ‘${R_2}$’ is given as,
${R_2} = R + r$
Thus, the equivalent resistance ‘${R_2}$’ is greater than the equivalent resistance of the group of resistances ‘R’. i.e. ${R_2} > R$.
Therefore, it can be concluded that,
${R_1} < R$
${R_2} > R$
Hence, both option (B) and (C) are correct answers.
Note:
A resistor is defined as the passive two-terminal electrical component which implements electrical resistance as a circuit element. Resistors are used to reduce the current flow and lower voltage levels within the circuits. Most circuits often contain more than one resistor in order to limit the flow of charges in a circuit.
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