I need help answering the last 3 please! EXAMPLE 18.1Four Resistors in Series (a

Question

I need help answering the last 3 please!

EXAMPLE 18.1Four Resistors in Series

(a) Four resistors connected in series. (b) The equivalent resistance of thecircuit in (a).

SOLUTION

(A) Find the equivalent resistance of the circuit.

Sum the resistances to find the equivalent resistance.

(B) Find the current in the circuit.

Apply Ohm's law to the equivalent resistor in figure b, solving for the current.

I =

=

= 0.33 A(C) Calculate the electric potential at point A.

Apply Ohm's law to the 2.0- resistor to find the voltage drop across it.

V = IR = (0.33 A)(2.0 ) = 0.66 V

To find the potential at A, subtract the voltage drop from the potential at the positive terminal.

VA = 6.0 V 0.66 V = 5.3 V

(D) Calculate the battery's internal resistance if the battery's emf is 6.2 V.

Write the following equation.

Solve for the internal resistance r and substitute values.

r =

=

= 0.6 (E) What fraction f of the battery's power is delivered to the load resistors?

Divide the power delivered to the load by the total power output.

f =

=

=

= 0.97

LEARN MORE

REMARKS A common misconception is that the current is "used up" and steadily declines as it progresses through a series of resistors. That would be a violation of conservation of charge. What is actually used up is the electric potential energy of the charge carriers, some of which is delivered to each resistor.

QUESTION Explain why the current in a real circuit very slowly decreases with time compared to its initial value.

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

A circuit consists of a battery with a closed circuit, a terminal voltage of 12.0 V, and 1.4-, 2.5-, 5.5-, and 9.5- resistors connected in series, oriented as in figure a above, with the battery in the bottom of the loop, positive terminal on the left, and resistors in increasing order, left to right, in the top of the loop.

Explanation / Answer

(a)Equivalent resistance= R1+R2+R3+R4= 1.4+2.5+5.5+9.5=18.9 Ohms

(b) Apply V=IR

12=I*18.9; I=0.6349= 634.9 mA

(c) P=VI

total power= 12*634.9*10-3 W = 7.62 W

(d) Current goes from left to right. So, voltage gets decreased when moving along left to right across a resistor.

potential decrement across 1.4 resistor(V1)=634.9*1.4 = 888.86 mV

potential decrement across 2.5 resistor(V2)=634.9*2.5= 1587.25 mV

potential of the mid point= 12-V1-V2= 9.524 V

(e)Let 'r' be the internal resistance

V=IR; 12.1= (R1+R2+R3+R4+r)*I;

I=12.1/(r+18.9);

voltage across r= 12.1-12=0.1;

Apply V=IR for 'r'

0.1= r*12.1/(r+18.9);

r=18.9/120 = 0.1575 Ohms= 157.5 mili Ohms

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.