Answer: V=IR
Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as
E = I(R + r) and by expanding, we have that E =IR + It
Where r is the internal resistance of the battery
I is the total current flowing in the circuit
R total load resistance in the circuit.
E is the total emf of the circuit.
The total emf is the sum of 2 separate voltages.
"IR" which is the terminal voltage and "Ir" which is the loss voltage.
The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.
The answer to this question is c
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The self-inductance of a coil will change by 8 times its original value by increasing its radius value by 2 and increasing the length of the coil by 2.
Self-Inductance: -
The definition of self-inductance is the induction of a voltage in a wire that carries current when the current in the wire is changing. In the instance of self-inductance, the circuit itself induces a voltage through the magnetic field produced by a changing current.
We know that the self-inductance of the coil is denoted by: -
L= µ *π*(r)^2*(N)^2*l
Where
L= Self-Inductance of the coil
µ= Magnetic Permeability Constant
r= Radius of the coil
l= Length of the coil
N= Number of turns of the coil
Here Self-inductance of the coil is directly proportional to the length of the coil and the square of the radius of the coil.
So,
On increasing the radius of the coil by a factor of 2 and the length of the coil by 2 the self-inductance of the coil increases by 8 times its original value.
Learn more about Self-Inductance here: -
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