Respuesta :
Let's call R the value of the resistance of the two resistors. In the first situation, the resistors are connected in parallel, so their equivalent resistance is given by:
[tex]\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}[/tex]
which means:
[tex]R_{eq}=\frac{R}{2}[/tex]
And calling I the current in the circuit, the voltage is given by Ohm's law:
[tex]V=IR_{eq}=\frac{IR}{2}[/tex]
In the second situation, the resistance of each resistor is doubled: R'=2R. So, the equivalent resistance in this case is given by
[tex]\frac{1}{R_{eq}}=\frac{1}{2R}+\frac{1}{2R}=\frac{2}{2R}=\frac{1}{R}[/tex]
which means
[tex]R_{eq}=R[/tex]
And the new voltage is given by:
[tex]V'=IR_{eq}=IR=2V[/tex]
which is twice the original voltage, so the voltage has doubled.
[tex]\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}[/tex]
which means:
[tex]R_{eq}=\frac{R}{2}[/tex]
And calling I the current in the circuit, the voltage is given by Ohm's law:
[tex]V=IR_{eq}=\frac{IR}{2}[/tex]
In the second situation, the resistance of each resistor is doubled: R'=2R. So, the equivalent resistance in this case is given by
[tex]\frac{1}{R_{eq}}=\frac{1}{2R}+\frac{1}{2R}=\frac{2}{2R}=\frac{1}{R}[/tex]
which means
[tex]R_{eq}=R[/tex]
And the new voltage is given by:
[tex]V'=IR_{eq}=IR=2V[/tex]
which is twice the original voltage, so the voltage has doubled.
Answer:
If the resistance in both resistors is doubled, then voltage gets doubled.
Explanation:
It is given that, A circuit consists of a battery and two resistors connected in parallel.
In parallel combination of resistors the potential difference remains the same while the current divides. The equivalent resistance is given by :
[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}[/tex]
Initial condition, Let R₁ = R₂ = R
So, [tex]\dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}[/tex]
[tex]{R_{eq}=\dfrac{R}{2}[/tex]
Using Ohm's law,
[tex]V=IR_{eq}=\dfrac{IR}{2}[/tex]...............(1)
Final condition, If the resistance in both resistors is doubled, and the current is kept constant.
So, [tex]\dfrac{1}{R'_{eq}}=\dfrac{1}{2R}+\dfrac{1}{2R}[/tex]
[tex]{R'_{eq}=R[/tex]
Voltage, [tex]V'=IR'_{eq}=IR[/tex]............(2)
From equation (1) and (2),
[tex]V'=2V[/tex]
So, the new voltage gets doubled. Hence, this is the required solution.
