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netineya [11]
3 years ago
13

A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re

sistors is 1.8
a. find (a) the emf of the battery and (b) the current through each resistor.
Physics
1 answer:
White raven [17]3 years ago
4 0
A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law. 

I= \frac{V}{R}

Where:
I = current (A)
V = Voltage (V) (emf)
R = Resitance (Ω)

Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
V = IR

However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

\frac{1}{R_{T}} =  \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}


\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059
\frac{1}{R_{T}} =0.0613

Get the reciprocal of both sides and divide:

R_{T} =  \frac{1}{0.0613} =16.32

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
V = IR
V = (1.8)(16.32)
V = 29.376V

The emf of the battery is 29.376V


B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
I= \frac{V}{R}
I= \frac{29.376}{65}
I= 0.45A

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
I= \frac{V}{R}
I= \frac{29.376}{25}
I= 1.18A
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
I= \frac{V}{R}
I= \frac{29.376}{170}
I= 0.17A

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.
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       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
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