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
1 answer:
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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Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
