Two resistors of 830 Ohm and 310 Ohm are connected in series to a 9.10-V battery

Question

Two resistors of 830 Ohm and 310 Ohm are connected in series to a 9.10-V battery. Find the current in this circuit, assuming the battery has no internal resistance. (0/5 submissions used) A A Capacitors accumulates 2.40e-5 C of charge when connected to a 9.0 volt (5.00) battery. What is its capacitance? 2.67-7 (1/5 submissions used) F Find the equivalent resistance of three resistors connected in parallel, whose resistances are 300 Ohm, 590 Ohm, and 330 Ohm. 12-gauge copper wire has a radius of 1.03 times 10^-3 m. (a) What is the resistance of 11.5 m of this wire at 20 degree C? (b) How much current will pass through the wire if a potential difference of 9.00 V is applied across it? (a) (2/5 submissions used) Ohm (b) 155 (1/5 submissions used) A

Explanation / Answer

7.3
resistors inseries are added algebria=cally, R1 + R2
so, let the current in the circuit be i
then V = iR( from ohms la)
now, V = 9.1 V battery ( from question )
and R = R1 + R2 = 830+310 = 1140 ohm
so i = V/R = 9.1/1140 = 0.007982 A

28.1.3 :
charge, q = 2.4*10^-5 C
battery, V = 9 V
now, q = cV
so, C = q/V = 2.4*10^-5/9 = 2.666 micro F

29. 11.2
Equivalent resistance of three resistors connected in series is
Reff = (1/R1 + 1/R2 + 1/R3)^-1
now R1 = 300 OHM
r2 = 590 ohm
R3 = 330 ohm
so, effective resistnace. R
R = (1/300 + 1/590 + 1/330)
R = 124.091 ohms
30., radius of wire, r = 10^-3 m
a) resistivity of cvopper, rho = 1.72*10^-8
length of wire, l = 11.5 m
resistance, R = rho*l/A = rho*11.5/pi*(10^-3)^2 = 1.72*10^-8 *11.5/PI*(10^-3) = 1.39081 OHM

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