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evablogger [386]
2 years ago
11

If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of beta will be :​

Mathematics
1 answer:
Tems11 [23]2 years ago
8 0
<h3>Given -:</h3>

change in emitter current, ∆l(E) = 4 mA

change in collector current,∆I(C) = 3.5 mA

<h3>To find -:</h3>

value of B

<h3>Solution :-</h3>

∆I(E) = ∆I(C) + ∆l(B), here ∆l(B) is the change in base current.

4 = 3.5 + ∆l(B)

∆I(B) = 0.5 mA

<h3>Now, B = ∆l (C) ÷ ∆l (B)</h3>

B = 3.5 ÷ 0.5

B = 7.0

<h3>so, the value of A = 7.0</h3>

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<u>Total Deductions</u> (from percentages) = 2641.2 + 596.4 + 1278 + 5538 = 10053.6

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3 years ago
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aleksandrvk [35]

Given that,

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ValentinkaMS [17]
 1)Rewrite the table:
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2) write the quotient of each number by the number before & notice the value:

49/70= 0.7
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24.01/34.3 =0.7
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3) Domain and Range of this function:
Last term = a₁.rⁿ⁻¹. let last term be y==> f(n) = y =70(0.7)ⁿ⁻¹
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Method 2

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