What is the current in a series circuit that has two resistors (4.0 ohms and 7.5 ohms) and a power source of 9.0 volts?

2 answers:

3 0

-- First, we have to decide how to handle the two resistors.

The effective resistance of resistors in series is the sum
of their individual resistances. That is, they act like a single
resistor, whose resistance is the sum of all of them.

So in this question, the 4.0 ohms and the 7.5 ohms act like a
single resistor of 11.5 ohms.

-- The current in the circuit is

   (the supply voltage) / (the total resistance)

   = (9.0 volts) / (11.5 ohms)

   = 0.783... Ampere (rounded)

hram777 [196]4 years ago

3 0

Answer choice 'D' for Plato users.

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Each of the two coils has 320 turns. The average radius of the coil is 6 cm. The distance from the center of one coil to the ele

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Answer:

Magnetic field, $B=2.39\times 10^{-3}\ T$

Explanation:

It is given that,

Number of turns, N = 320

Radius of the coil, r = 6 cm = 0.06 m

The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m

Current flowing through the coils, I = 0.5 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

$B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}$

B = 0.00239 T

$B=2.39\times 10^{-3}\ T$

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is $2.39\times 10^{-3}\ T$ . Hence, this is the required solution.