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Given a 64.0 V battery and 30.0 Ω and 88.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W (b) Repeat when the resistances are in parallel. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W

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oladayoademosu

Answer:

a. i. 0.542 A ii. 8.813 W iii. 0.542 A iv. 25.85 W

b. i. 2.13 A ii. 136.53 W iii. 0.727 A iv. 46.55 W

Explanation:

a. Find the current (in A) and power (in W) for each when connected in series.

Since the resistors are connected in series, their combined resistance is R = R₁ + R₂ where R₁ = 30.0 Ω and R₂ = 88.0 Ω.

So, substituting the values of the variables into the equation, we have

R = R₁ + R₂

R =  30.0 Ω +  88.0 Ω

R =  118.0 Ω

Since from Ohm's law, V = IR where V = voltage across circuit = battery voltage = 64.0 V, I = current in circuit and R = total resistance of circuit = 118.0 Ω

So, I = V/R = 64.0V/118.0 Ω = 0.542 A

Since the resistors are in series, the same current flows through them

i. Current in 30.0 Ω

Current in 30.0 Ω is I = 0.542 A since the resistors are in series.

ii Power in the 30.0 Ω

The power in the 30.0 Ω is P₁ = I²R₁ where I = current = 0.542 A and R₁ = resistance = 30.0 Ω

So, P₁ = I²R₁

= (0.542 A)² × 30.0 Ω

= 0.293764  A² × 30.0 Ω

= 8.8129 W

≅ 8.813 W

iii. Current in 88.0 Ω

Current in 88.0 Ω is I = 0.542 A since the resistors are in series.

iv. Power in the 88.0 Ω

The power in the 88.0 Ω is P = I²R₂ where I = current = 0.542 A and R₂ = resistance = 88.0 Ω

So, P₂ = I²R₂

= (0.542 A)² × 88.0 Ω

= 0.293764  A² × 88.0 Ω

= 25.8512 W

≅ 25.85 W

(b) Repeat when the resistances are in parallel.

Since the resistors are connected in parallel, the same voltage is applied across them.

i. Current in 30.0 Ω

Using Ohm's law, V = I₁R₁ where V = voltage = 64.0 V, I₁ = current in 30.0 Ω resistor and R₁ = resistance = 30.0 Ω

So, I₁ = V/R₁ = 64.0 V/30.0 Ω = 2.13 A

ii Power in the 30.0 Ω

The power in the 30.0 Ω resistor is P₁ = V²/R₁ where V = voltage across resistor = 64.0 V and R₁ = resistance = 30.0 Ω

So, P₁ = V²/R₁

P₁ = (64.0 V)²/30.0 Ω

P₁ = 4096 V²/30.0 Ω

P₁ = 136.53 W

iii. Current in 88.0 Ω

Using Ohm's law, V = I₂R₂ where V = voltage = 64.0 V, I₂ = current in 88.0 Ω resistor and R₂ = resistance = 88.0 Ω

So, I₂ = V/R₂ = 64.0 V/88.0 Ω = 0.727 A

iv. Power in the 88.0 Ω

The power in the 30.0 Ω resistor is P₂ = V²/R₂ where V = voltage across resistor = 64.0 V and R₂ = resistance = 88.0 Ω

So, P₂ = V²/R₂

P₂ = (64.0 V)²/88.0 Ω

P₂ = 4096 V²/88.0 Ω

P₂ = 46.55 W

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