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Alenkasestr [34]
3 years ago
8

Two 20 ohm resistors are connected in parallel and two 10 ohm resistors are connected in parallel. If these two combinations are

connected in series the equivalent resistance of the combination is: a) 20 ohm b) 10 ohm c) 15 Ohm d) 30 Ohm

Physics
1 answer:
geniusboy [140]3 years ago
5 0

C. 15Ω.

The circuit diagram is shown in the first image attached. To solve this problem we have to use the equivalent resistor conected in parallel Req = (R1*R2)/(R1+R2).

Calculating the equivalent resistor in the 20Ω resistor parallel arrangement:

RAeq = [(20*20)/(20+20)] Ω

RAeq = 400/40 Ω

RAeq = 10Ω

Calculating the equivalent resistor in the 10Ω resistor parallel arrangement:

RBeq = [(10*10)/(10+10)]Ω

RBeq = 100/20 Ω

RBeq = 5Ω

Both equivalent resistors are in series. So, in order to calculate the equivalent resistance of the combination:

Rtotal = RAeq + RBeq

Rtotal = 10Ω + 5Ω

Rtotal = 15Ω

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Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
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Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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