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2 years ago

12

in a circuit, a 10 resistor is connected in series to a parallel group containing a 60 resistor and a 5 resistor. what is the to

tal resistance in this circuit?

2 answers:

guajiro [1.7K]2 years ago

7 0

60 ohm and 5 ohm resistors are in parallel and their parallel combination is given as

$R_{parallel}$ = Product of resistances / (Sum of resistance)

$R_{parallel}$ = 60 x 5 /(60 + 5)

$R_{parallel}$ = 4.62 ohm

The parallel combination of resistances " $R_{parallel}$ " and 10 ohm resistor are in series and their equivalent resistance is given as

$R_{total}$ = $R_{parallel}$ + 10

$R_{total}$ = 4.62 + 10

$R_{total}$ = 14.62 ohm

natka813 [3]2 years ago

5 0

The 60 and the 5 in parallel have an equivalent resistance of 4.615 ohms. (rounded). The 10 in series makes it 14.615...

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2 years ago

At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

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Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

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=> $A_2 = 4 A_1$

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=> $v_2 = \frac{3.57}{4}$

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Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

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substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

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Intertidal zone:brainly.com/question/1346004

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