Answer:
Explanation:
Area of electrodes, A = 2 cm x 2 cm = 4 cm²
Separation between electrodes, d = 1 mm
Voltage, V = 9 V
(a)
Let C is the capacitance between the electrodes
C = 3.54 x 10^-12 F
Let q be the charge on each of the electrode
q = C x V
q = 3.54 x 10^-12 x 9 = 3.2 x 10^-11 C
(b)
As, the battery is disconnected the charge on the electrodes remains same.
(c)
As the battery is connected the voltage is same.
capacitance is change.
As the distance is doubled, the capacitance becomes half and the charge is also halved. q' = q/2 = 1.6 x 10^-11 C
18 degree is equal to 64.4 fahrenheit
Answer:
351 ohm
720 ohm
Explanation:
When c and d are open:
Terminals c and d are open.. If you redraw the circuit as below, you can see that the two resistors in the first column are in parallel as, they are connected together at both pairs of terminals (due to the short).
Hence, we have a pair of parallel resistors:
Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms
Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms
Now these two sets are in series with another Hence,
Req = Req1 + Req2 = 216 + 135 = 351 ohms
Answer: 351 ohms
When c and d are shorted:
The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.
Both of these resistor lie in a single path placing the resistors in series to one another, hence
Req = R3 + R1 = 180 + 540 = 720 ohms
Answer:720 ohms
