Respuesta :
Answer:
A) [tex]8.99[/tex] V
B) [tex]2.5[/tex] Ohm
Explanation:
Two resistors are connected in parallel circuit. Thus, the equivalent resistance across the circuit is equal to
[tex]\frac{I}{R_e} = \frac{I}{R_1} + \frac{I}{R_2}\\[/tex]
Substituting the values of [tex]R_1[/tex] and [tex]R_2[/tex], we get
[tex]\frac{1}{R_e} = \frac{1}{45} + \frac{1}{75}\\\frac{1}{R_e} = 0.03555\\R_e = 28.125[/tex]
Let the internal resistance of the battery be represented by "r"
As per Ohm's law
[tex]Emf = IR[/tex]
Where I is the current and R is the resistance
When [tex]45.0[/tex]-V resistor is disconnected, the current from the battery drops to [tex]0.116[/tex] A.
Thus,
[tex]Emf = 0.294 (28.125 + r)[/tex] -------- Eq (1)
[tex]Emf = 0.116 (75 + r)[/tex] -------- Eq (2)
Solving equation 1 and 2 , we get
[tex]Emf = 8.99[/tex] V
Substituting the value of Emf in equation 1, we get -
[tex]r = 2.5[/tex] ohm
Answer:
The emf is 2.42 V.
The internal resistance of the battery is 8.98 V.
Explanation:
Given that,
Resistance [tex]R_{1}=75.0 \Omega[/tex]
Resistance [tex]R_{2}=45.0\ Omega[/tex]
Current = 0.294 A
Drop current = 0.116 A
We need to calculate the equivalent resistance
Using formula of equivalent resistance
[tex]\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{1}{R}=\dfrac{1}{75}+\dfrac{1}{45}[/tex]
[tex]R=28.125\ \Omega[/tex]
We need to calculate the voltage
Using formula of voltage
[tex]V=IR[/tex]
Put the value into the formula
[tex]V=0.294\times28.125[/tex]
[tex]V=8.27\ volt[/tex]
We need to calculate the voltage after removed the 45.0 ohm resistance
Using formula of voltage
[tex]V=IR[/tex]
Put the value into the formula
[tex]V=0.116\times75.0[/tex]
[tex]V=8.7\ volt[/tex]
EMF and internal resistance of the battery is constant
We need to calculate the internal resistance
Using formula of emf
[tex]E=V+ir[/tex]
Put the value into the formula
[tex]E=8.27+0.294r[/tex]...(I)
[tex]E=8.7+0.116r[/tex]...(II)
Subtract equation (II) from equation (I)
[tex]0=0.43-0.178r[/tex]
[tex]r=\dfrac{0.43}{0.178}[/tex]
[tex]r=2.42\ \Omega[/tex]
Put the value of r in equation (I)
[tex]E=8.27+0.294\times2.42[/tex]
[tex]E=8.98\ V[/tex]
Hence, The emf is 2.42 V.
The internal resistance of the battery is 8.98 V.
